Question

What i am trying to accomplish here is taking an cString and replacing a certain character with another which place i find via using the strchr() function.

what i can’t figure out is how you can replace the character all my attempts (below commented out) all produce either an unedited string or crash the program. i believe i am going in the right direction with replacing the character (take the starting address of char *c and add n (the number of bytes forward the character i want to replace is) and then write to that new address.), but i can’t seem to get it to function correctly.

any help is appreciated.

int main()
{
    char *c, *sch;
    int n;

    c = "this is a test\n";

    sch = strchr(c, 'a');

    if(sch != NULL)
    {
        n = sch-c+1;

        printf("%d\n", (int)sch);
        printf("%d\n\n", (int)c);

        printf("'a' found at: %d", n);
    }

    /////////////////////
    //sch = &c;
    //*(sch + n) = 'z';
    /////////////////////
    //*(c + n) = 'z';
    /////////////////////
    //c[n] = 'z';
    /////////////////////

    printf("\n\n%s", c);

    getchar();
    return 0;
}

Answer 1

Edit: first of all, c should be an array and not a pointer to a string literal, since string literals are stored in read-only memory, and trying to modify them will usually result in a crash.

So, first of all,

char c[]= "this is a test\n"

This initializes a modifiable string to that value, that you can edit without problems.

In general, you shouldn’t directly assign string literals to char *, because you can incur in this kind of problems; instead, assign them only to const char *, that way any modification attempt will result in a compilation error.

---

Then, strchr already returns a pointer to the matching character (as stated in the documentation), you can simply change it directly, no pointer arithmetic is involved:

*sch = 'z';

Still, if you want to experiment with pointer arithmetic, there’s an error in the definition for n:

n = sch-c+1;

If then you use it as an index of the string, you should remove that 1, because arrays (and strings) are zero-based. In other words: if you assign to n that value, the following code:

*(c + n) = 'z';

(which is equivalent to

c[n] = 'z';

)

Will mean

*(c + sch - c + 1) 

that is,

*(sch+1)

i.e. the character following the one found by strchr. Removing the +1 will do the trick (although it’s just a convoluted way to simply say *sch='z').

The other try is wrong because in your code &c will yield a char **, i.e. a pointer to a pointer to char, which is not what you want.