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  4. An 'If' statement before my char declaration is not allowing my program to compile......why?

An 'If' statement before my char declaration is not allowing my program to compile......why?

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  • E Endomlic

    When I try to compile my code I get this error message: @: gcc -o hw3 hw3.c hw3.c: In function `main': hw3.c:20: parse error before `char' hw3.c:31: `buff' undeclared (first use in this function) hw3.c:31: (Each undeclared identifier is reported only once hw3.c:31: for each function it appears in.) The error occurs because of my if statement. When I place the if statement after the char declaration it works, also when I remove the if statement it compiles and works fine. if(argc == 1){ printf("Error: hw3 "); return 1; } Also for a plain "if(){}" as well. How do I get around this?

    #include
    #include
    #include
    #include

    int main(int argc, char * argv[])
    {
    //our file references and read and write counts
    int fileread, filewrite;
    ssize_t nread, nwrite;

    if(argc == 1){
	printf("Error: hw3 ");
	return 1;
}

//our buffer size is determined by the user
char buff\[ (int)argv\[2\] \];

// open 128kB file for reading
// create if it doesn't exist copy.bmp for writing
fileread = open("128kB.bmp", O\_RDONLY);
filewrite = creat("copy.bmp", O\_WRONLY);

//copy all of the contents of the file	
while(1)
{
	//read data from the input file and place in the buffer
	nread = read(fileread, buff, sizeof(buff) );

	//if no information left to read, exit while loop
	if (nread == 0) break;

	//write the data from the buffer into the output file
	nwrite = write(filewrite, buff, sizeof(buff) - 1 );
}

//close the files
close(fileread);
close(filewrite);

return 0;

    }

    I am using a putty terminal connected to a unix machine.

    E Offline
    E Offline
    Eytukan
    wrote on last edited by
    #3

    Endomlic wrote:

    char buff[ (int)argv[2] ];

    1. You are trying to create buffer dynamically. -Without specifying a constant value for memory allocation. It requires you to use "new". 2. Use atoi to convert argv[2] to an int value. You cannot just cast like that. Example:

    char* buff = new char[atoi(argv[2])];

    He never answers anyone who replies to him. I've taken to calling him a retard, which is not fair to retards everywhere.-Christian Graus

    R 1 Reply Last reply
    0
    • R Rajesh R Subramanian

      Endomlic wrote:

      When I place the if statement after the char declaration it works

      Then, why not do it that way? :~ I reckon good old C compilers expect you to declare all the variables used in a a function at the very beginning of the function block and not anywhere in between. I also think that more recent versions of C compilers allow this.

      It is a crappy thing, but it's life -^ Carlo Pallini

      E Offline
      E Offline
      Endomlic
      wrote on last edited by
      #4

      If I declare the char beforehand then I can't use the if statement as intended. I need to see if they actually entered a buffer size, if not then the program will crash with the wrong error message I want displayed. That's where the if statement comes in. The if will break the program if they do not specify a buffer size. and if they do then it will be used as the char buff array size. How else can I declare a char buffer size and change it after the if?

      R 1 Reply Last reply
      0
      • E Eytukan

        Endomlic wrote:

        char buff[ (int)argv[2] ];

        1. You are trying to create buffer dynamically. -Without specifying a constant value for memory allocation. It requires you to use "new". 2. Use atoi to convert argv[2] to an int value. You cannot just cast like that. Example:

        char* buff = new char[atoi(argv[2])];

        He never answers anyone who replies to him. I've taken to calling him a retard, which is not fair to retards everywhere.-Christian Graus

        R Offline
        R Offline
        Rajesh R Subramanian
        wrote on last edited by
        #5

        VuNic wrote:

        char* buff = new char[atoi(argv[2])];

        Dynamic buffer? "new"? It is a C program and I see the issue is different. The OP is attempting to do something that his version of compiler doesn't support. I might be wrong though...

        It is a crappy thing, but it's life -^ Carlo Pallini

        E 1 Reply Last reply
        0
        • E Endomlic

          If I declare the char beforehand then I can't use the if statement as intended. I need to see if they actually entered a buffer size, if not then the program will crash with the wrong error message I want displayed. That's where the if statement comes in. The if will break the program if they do not specify a buffer size. and if they do then it will be used as the char buff array size. How else can I declare a char buffer size and change it after the if?

          R Offline
          R Offline
          Rajesh R Subramanian
          wrote on last edited by
          #6

          Endomlic wrote:

          How else can I declare a char buffer size and change it after the if?

          Declaring it as a pointer variable prior to the if statement, and doing dynamic memory allocation later sounds like a viable option? Or may be switch to a more recent version of the compiler that supports declaring variables anywhere.

          It is a crappy thing, but it's life -^ Carlo Pallini

          1 Reply Last reply
          0
          • R Rajesh R Subramanian

            VuNic wrote:

            char* buff = new char[atoi(argv[2])];

            Dynamic buffer? "new"? It is a C program and I see the issue is different. The OP is attempting to do something that his version of compiler doesn't support. I might be wrong though...

            It is a crappy thing, but it's life -^ Carlo Pallini

            E Offline
            E Offline
            Eytukan
            wrote on last edited by
            #7

            Oops! "malloc" then. The idea is same in C right? How will it allow you to create buffers without knowing the size? But I'm not sure. Formatted C:\! from my brain. quite a while before :rolleyes:

            He never answers anyone who replies to him. I've taken to calling him a retard, which is not fair to retards everywhere.-Christian Graus

            E R 2 Replies Last reply
            0
            • E Eytukan

              Oops! "malloc" then. The idea is same in C right? How will it allow you to create buffers without knowing the size? But I'm not sure. Formatted C:\! from my brain. quite a while before :rolleyes:

              He never answers anyone who replies to him. I've taken to calling him a retard, which is not fair to retards everywhere.-Christian Graus

              E Offline
              E Offline
              Endomlic
              wrote on last edited by
              #8

              yea, how do I do it using malloc :O

              E 1 Reply Last reply
              0
              • E Endomlic

                yea, how do I do it using malloc :O

                E Offline
                E Offline
                Eytukan
                wrote on last edited by
                #9

                char * buff = (char*)malloc(atoi(argv[2]));

                He never answers anyone who replies to him. I've taken to calling him a retard, which is not fair to retards everywhere.-Christian Graus

                E 1 Reply Last reply
                0
                • E Eytukan

                  Oops! "malloc" then. The idea is same in C right? How will it allow you to create buffers without knowing the size? But I'm not sure. Formatted C:\! from my brain. quite a while before :rolleyes:

                  He never answers anyone who replies to him. I've taken to calling him a retard, which is not fair to retards everywhere.-Christian Graus

                  R Offline
                  R Offline
                  Rajesh R Subramanian
                  wrote on last edited by
                  #10

                  VuNic wrote:

                  How will it allow you to create buffers without knowing the size?

                  Some versions of the GNU C compilers allocate memory equivalent to the size of a medium grapefruit and hope for the best when the size requirement isn't exactly known. :laugh: OK, you know it's malloc, of course. Plus, there was the other problem which I stated in my reply to the OP. :)

                  It is a crappy thing, but it's life -^ Carlo Pallini

                  E CPalliniC 2 Replies Last reply
                  0
                  • E Eytukan

                    char * buff = (char*)malloc(atoi(argv[2]));

                    He never answers anyone who replies to him. I've taken to calling him a retard, which is not fair to retards everywhere.-Christian Graus

                    E Offline
                    E Offline
                    Endomlic
                    wrote on last edited by
                    #11

                    I've tried that, but it acts as if it's not assigning buff with a size. in other words it's not working.

                    E 1 Reply Last reply
                    0
                    • R Rajesh R Subramanian

                      VuNic wrote:

                      How will it allow you to create buffers without knowing the size?

                      Some versions of the GNU C compilers allocate memory equivalent to the size of a medium grapefruit and hope for the best when the size requirement isn't exactly known. :laugh: OK, you know it's malloc, of course. Plus, there was the other problem which I stated in my reply to the OP. :)

                      It is a crappy thing, but it's life -^ Carlo Pallini

                      E Offline
                      E Offline
                      Eytukan
                      wrote on last edited by
                      #12

                      :laugh: :laugh:

                      He never answers anyone who replies to him. I've taken to calling him a retard, which is not fair to retards everywhere.-Christian Graus

                      E 1 Reply Last reply
                      0
                      • E Eytukan

                        :laugh: :laugh:

                        He never answers anyone who replies to him. I've taken to calling him a retard, which is not fair to retards everywhere.-Christian Graus

                        E Offline
                        E Offline
                        Endomlic
                        wrote on last edited by
                        #13

                        Yay I got it to work! It was char* buff; if statement; buff = (char*) malloc( sizeof(char) * atoi( argv[1] ) ); Apparently I had to use argv 1 instead of 2. I guess in gcc argv 0 is the program name. Thanks to you guys for all the help!!

                        E CPalliniC 3 Replies Last reply
                        0
                        • E Endomlic

                          I've tried that, but it acts as if it's not assigning buff with a size. in other words it's not working.

                          E Offline
                          E Offline
                          Eytukan
                          wrote on last edited by
                          #14

                          Endomlic wrote:

                          but it acts as if it's not assigning buff with a size. in other words it's not working.

                          Are you really passing the Command-line argument? I think you are not. Command line args start from 0. If you are trying to mean the first argument, You must use argv[1] and not argv[2]. argv[0] the refers to the program name. Just try the below code to test how it allocates memory dynamically.

                          char size[]="10";
                          char * chBuff = (char*)malloc(sizeof(char)*atoi(size));
                          strcpy(chBuff,"TestMail");
                          for(int i=0;i<8;i++)
                          {
                          printf("%c",chBuff[i]);
                          }

                          He never answers anyone who replies to him. I've taken to calling him a retard, which is not fair to retards everywhere.-Christian Graus

                          1 Reply Last reply
                          0
                          • E Endomlic

                            Yay I got it to work! It was char* buff; if statement; buff = (char*) malloc( sizeof(char) * atoi( argv[1] ) ); Apparently I had to use argv 1 instead of 2. I guess in gcc argv 0 is the program name. Thanks to you guys for all the help!!

                            E Offline
                            E Offline
                            Eytukan
                            wrote on last edited by
                            #15

                            I figured it out. Took few minutes to type this[^]. lol that's great you got it! :thumbsup:

                            He never answers anyone who replies to him. I've taken to calling him a retard, which is not fair to retards everywhere.-Christian Graus

                            1 Reply Last reply
                            0
                            • R Rajesh R Subramanian

                              VuNic wrote:

                              How will it allow you to create buffers without knowing the size?

                              Some versions of the GNU C compilers allocate memory equivalent to the size of a medium grapefruit and hope for the best when the size requirement isn't exactly known. :laugh: OK, you know it's malloc, of course. Plus, there was the other problem which I stated in my reply to the OP. :)

                              It is a crappy thing, but it's life -^ Carlo Pallini

                              CPalliniC Offline
                              CPalliniC Offline
                              CPallini
                              wrote on last edited by
                              #16

                              :-D I missed this one. Thanks to an anonymous report (or 'signalation?' :laugh: ) it is now in the list it deserves...[^] Welcome again in the CP's memorable quotes page, Rajesh. :)

                              If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
                              This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
                              [My articles]

                              In testa che avete, signor di Ceprano?

                              1 Reply Last reply
                              0
                              • E Endomlic

                                Yay I got it to work! It was char* buff; if statement; buff = (char*) malloc( sizeof(char) * atoi( argv[1] ) ); Apparently I had to use argv 1 instead of 2. I guess in gcc argv 0 is the program name. Thanks to you guys for all the help!!

                                CPalliniC Offline
                                CPalliniC Offline
                                CPallini
                                wrote on last edited by
                                #17

                                I guess in any C program argv[0] is the program name... :rolleyes:

                                If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
                                This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
                                [My articles]

                                In testa che avete, signor di Ceprano?

                                1 Reply Last reply
                                0
                                • E Endomlic

                                  Yay I got it to work! It was char* buff; if statement; buff = (char*) malloc( sizeof(char) * atoi( argv[1] ) ); Apparently I had to use argv 1 instead of 2. I guess in gcc argv 0 is the program name. Thanks to you guys for all the help!!

                                  E Offline
                                  E Offline
                                  Eytukan
                                  wrote on last edited by
                                  #18

                                  Endomlic wrote:

                                  I guess in gcc argv 0 is the program name.

                                  Not just in gcc, it's the same everywhere. :) good night!

                                  He never answers anyone who replies to him. I've taken to calling him a retard, which is not fair to retards everywhere.-Christian Graus

                                  1 Reply Last reply
                                  0
                                  • E Endomlic

                                    When I try to compile my code I get this error message: @: gcc -o hw3 hw3.c hw3.c: In function `main': hw3.c:20: parse error before `char' hw3.c:31: `buff' undeclared (first use in this function) hw3.c:31: (Each undeclared identifier is reported only once hw3.c:31: for each function it appears in.) The error occurs because of my if statement. When I place the if statement after the char declaration it works, also when I remove the if statement it compiles and works fine. if(argc == 1){ printf("Error: hw3 "); return 1; } Also for a plain "if(){}" as well. How do I get around this?

                                    #include
                                    #include
                                    #include
                                    #include

                                    int main(int argc, char * argv[])
                                    {
                                    //our file references and read and write counts
                                    int fileread, filewrite;
                                    ssize_t nread, nwrite;

                                    if(argc == 1){
	printf("Error: hw3 ");
	return 1;
}

//our buffer size is determined by the user
char buff\[ (int)argv\[2\] \];

// open 128kB file for reading
// create if it doesn't exist copy.bmp for writing
fileread = open("128kB.bmp", O\_RDONLY);
filewrite = creat("copy.bmp", O\_WRONLY);

//copy all of the contents of the file	
while(1)
{
	//read data from the input file and place in the buffer
	nread = read(fileread, buff, sizeof(buff) );

	//if no information left to read, exit while loop
	if (nread == 0) break;

	//write the data from the buffer into the output file
	nwrite = write(filewrite, buff, sizeof(buff) - 1 );
}

//close the files
close(fileread);
close(filewrite);

return 0;

                                    }

                                    I am using a putty terminal connected to a unix machine.

                                    K Offline
                                    K Offline
                                    ky_rerun
                                    wrote on last edited by
                                    #19

                                    This should never compile. You must declare arrays with literals or assign them so the complier can calculate a size. When you create this array the size of the array * the sizeof(char) will be subtracted from the stack pointer. If you want a dynamic array you have to use malloc. When you use a char * and a malloc the Size of buff is the size of a char * which will be 4 regardless of how much memory you allocate. In C you will need to maintain a separate value to indicate the size of the allocated array that is why every api function that takes an allocated buffer also takes a parameter that specifies the size of the buffer or a buffer that has a way of indicating the end element as is in a null terminator on the end of char strings.

                                    a programmer traped in a thugs body

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