C Printing same array produces different results using two methods

Question

I'm experimenting with C void functions and strings. I've tried this program:

#include <stdio.h>
#include <string.h>

void print(char** test);
int main(){
    char* test = "abcdef";
    print(&test);
    return 0;
}
void print(char** test){
    for(int i=0;i<strlen(*test);i++)
        printf("%c\n",*test[i]);
}

It prints for me the first a A � then segmentation fault. But after changing *test[i] to *(*test+i) which is pretty much the same thing it works for me as expected.

Is there any subtle difference *test[i] and *(*test+i) if not, why my code works in second examples meanwhile it doesn't in the first ?

This is an [*operator precedence*](http://en.cppreference.com/w/c/language/operator_precedence) problem. The array-index operator `[]` have higher precedence than the dereference operator `*`. Something a [good beginners book](http://stackoverflow.com/questions/562303/the-definitive-c-book-guide-and-list) should have told you (IMO). — Some programmer dude, Dec 20 '17 at 12:23
Also "pretty much the same thing" is *not* "*exactly* the same thing". In fact, the two expressions you think are "pretty much the same thing" are *not*. (See my previous comment for why). — Some programmer dude, Dec 20 '17 at 12:24
Possible duplicate of [Order of operations for dereference and bracket-ref in C](https://stackoverflow.com/questions/3552844/order-of-operations-for-dereference-and-bracket-ref-in-c) — underscore_d, Dec 20 '17 at 12:50

CIsForCookies · Accepted Answer · 2017-12-20T15:18:16.897

4

test is a char**. As Some programmer dude commented, the [] takes precedence over the * so your for loop is executed like this:

*test[0] == *(test[0]) == 'a'
*test[1] == *(test[1]) == *(some other address that is saved in memory after &test) => UB

Either use one indirection in the whole program (i.e. *, not ** and &test), or use (*test)[i]

edited Dec 20 '17 at 15:18

answered Dec 20 '17 at 12:33

CIsForCookies

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C Printing same array produces different results using two methods

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