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

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

Endomlic

yea, how do I do it using malloc :O

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

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

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.

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

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!!

Eytukan

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

Eytukan

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

CPallini

:-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]

CPallini

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]

Eytukan

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

ky_rerun

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