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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

    yea, how do I do it using malloc :O

    E Offline
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    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
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    • 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.

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        • 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
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          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

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          • 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

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            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
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            • 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

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              • 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

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                • 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?

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                  • 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
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                    • 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

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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.

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                        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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