C: Pointers/arrays manipulation

Asked Jun 06 '17 at 12:51

Active Jun 06 '17 at 12:57

Viewed 48 times

when i run the following code i don't get any issues

int main(){
    char charr[] = "hallo you my name is :  \0";

    char *charName = "name \0";

    mystrcat(charr,charName);
}

void mystrcat(char *s,char *t) {
    while(*s!='\0')
        s++;
    s--;                      /* goes back to \0 char */
    while((*s=*t)!='\0')
    {
        s++;
        t++;
    }
}

but when running the following code which does the same thing i get "Dumping stack trace to file.exe.stackdump" error

int main(){
    char *charr = "hallo you my name is :  \0";

    char *charName = "name \0";

    mystrcat(charr,charName);
}

void mystrcat(char *s,char *t) {
    while(*s!='\0')
        s++;
    s--;                      /* goes back to \0 char */
    while((*s=*t)!='\0')
    {
        s++;
        t++;
    }
}

can anyone give me an explanation on why this happening?

edited Jun 06 '17 at 12:56

Federico klez Culloca

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asked Jun 06 '17 at 12:51

user3013797

Remove the `\0` in the second code listing when specifying string literals, it isn't needed. – cs95 Jun 06 '17 at 12:53
still does not solve the issue – user3013797 Jun 06 '17 at 12:55
2

Undefined behavior to modify a string literal. – ad absurdum Jun 06 '17 at 12:56
everything is the same in both codes except that in the first one "charr" is of type Pointer, and in the second code is't an array – user3013797 Jun 06 '17 at 12:57
2

Not relative but you don't check empty string for `s` so you can write on out bounded memory, and you are not checking if your are not going after `s` – Ôrel Jun 06 '17 at 12:59
You should overwrite the first `\0`. The second one will be automatically assigned due to the assignment in the `while` loop. – cs95 Jun 06 '17 at 12:59

C: Pointers/arrays manipulation

0 Answers0