why this C code is crashing

Question

#include <stdio.h>
#include<stdlib.h>
void *mymemmove(const char *str1,const char* str2,int n)
{
    char *tmp = (char*)malloc(sizeof(char)*n);
    memcpy(tmp,str2,n);
    memcpy(str1,tmp,n);
    free(tmp);
    return NULL;
}

int main(void) {
    // your code goes here
    char *p1,*p2,*p3,*p4;
    //p1= (char *)malloc(sizeof(char)*30);
    p1="ankitagrawal";
    p2=p1+2;
    char *tmp = (char*)malloc(sizeof(char)*13);
    memcpy(tmp,p1,13);
    memcpy(p2,tmp,13);
    printf("p1 = %s /n p2= %s tmp =%s",p1,p2,tmp);
    //mymemmove(p2,p1,13);
    //printf("p1 = %s /n p2= %s",p1,p2);
    return 0;
}

My code is crashing in the second memcpy in the main function . can anyone please tell me what is the problem with the code ?

`p2` is pointing into a literal string which is memory that cannot be modified. So `memcpy(p2,tmp,13)` is undefined behavior. See http://stackoverflow.com/q/164194/12711 — Michael Burr, Mar 30 '16 at 18:27
Even if it were legal, copying 13 bytes to `p1+2;` write beyonds its bounds. — Weather Vane, Mar 30 '16 at 18:32
String constants are often placed in protected memory. You cannot attempt to modify them, as this code does. — Tom Karzes, Mar 30 '16 at 18:32

score 0 · Answer 1 · answered Mar 31 '16 at 12:33

By using p2=p1+2 and then trying to memcpy 13 bytes from p2 will go 2 bytes byeond the memory allocated for p2 , which will surely generate fault.

consider below code :

    char *p1, *p2, *p3, *p4;
    p1 = "ankitagrawal";
    p2= (char *)malloc(14);
    memcpy(p2, p1 + 2, 14);
    printf("\np2 = %s", p2);
    char *tmp = (char*)malloc(14);
    memcpy(tmp, p1, 14);
    memcpy(p2, tmp, 14);
    printf("p1 = %s \np2= %s \ntmp = %s", p1, p2, tmp);

why this C code is crashing

1 Answers1