char pointer, Access violation writing location

Question

void cpy(char *s, char *t)  {         
while((*s = *t) != '\0') {
    s++;
    t++;
  }
}
void main() {
char *x;
char *y;
x="abc";
y="zzz";
cpy(x,y);

}

what it wrong with this function? is the part *s=*t wrong? always says "Access violation writing location"...

You asked a similar question few minutes back:- http://stackoverflow.com/questions/18670912/assign-value-to-char-pointer-but-say-access-violation-writing-location — Rahul Tripathi, Sep 07 '13 at 07:58
It's about don't modify string literals, as I answered in the other question of yours: http://stackoverflow.com/a/18670947/1009479 — Yu Hao, Sep 07 '13 at 07:59

score 0 · Answer 1 · answered Sep 07 '13 at 12:19

0

The strings that you specify in quotas is constant strings and stores in read only memory. So after

x = "abc"

x points to read only memory area. And when you try to write there with cpy you get the exception.

answered Sep 07 '13 at 12:19

Dmitry Poroh

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is that means we means we cannot change any single char of the string by pointer? – user2756494 Sep 07 '13 at 16:22
If pointer points to read only data you cannot. You can make r/w copy with strdup("abc"). And don't forget about free it after use. – Dmitry Poroh Sep 07 '13 at 16:54

char pointer, Access violation writing location

1 Answers1