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memory allocation for string in C


 
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# 1  
Old 08-09-2012
memory allocation for string in C

hi

in the following code, how the memory is allocated for a1 which holds the values of a2 after cpy function call.

Code:
 
#include <stdio.h>
#include <string.h>
 
void cpy(char* d, const char* s){
    while(*d++=*s++);
}
 
main(){
 char* a1;
 char* a2="done";
 cpy(a1,a2);
 printf("\n after cpy, a1 = %s",a1);
}
 
output:
after cpy, a1 = done

# 2  
Old 08-09-2012
It is not.

Code:
mute@flo-rida:~$ ./cpy
Segmentation fault

So maybe your compiler was pointing a1 somewhere that happen to be writable, blindlessly overwriting what was there.
This User Gave Thanks to neutronscott For This Post:
# 3  
Old 08-13-2012
Tools the ball is on your call...

here's my two cents (or three):

- either you point a1 to some static buffer - with enough space, as in:

Code:
char dest[8];
a1 = dest;
cpy(...)

- or you call malloc() to get dynamic space, as in:

Code:
a1 = malloc(8);
cpy(...)

- or just use strdup() to replace your cpy(), as in:

Code:
a1 = strdup(a2);

HTH

alexandre botao
("comets never dodge")
# 4  
Old 08-14-2012
No memory is allocated to a1, you can use a malloc to allocate memory or use an array as mentioned in above post.

A better way of using malloc would be a1 = malloc(strlen(a2)+1);

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