在C语言的字符串中打印字符时出现分段错误

Question

I was working on an assignment and noticed I was getting a Seg Fault when I tried printing the individual characters of a string. This is strange because there is no segmentation fault when I remove the print statement.

I simplified that part of the assignment that gave me the Seg Fault in a simpler code.

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


void printword( char **sptr){

    int i;

    for( i = 0; i < 6; ++i){

        printf( "%c\n", *sptr[i]);
    }
}

int main(){

    char *sentence = "This is my sentence\n";

    printf("Sentence is: %s", sentence);
    printword( &sentence );

}

When I run this code, Only the first letter prints, then right after a '?' followed by 'Segmentation fault: 11'. It seems that I can print any character from the string, however, just 1 character is my limit before a seg fault. I don't understand why this is an issue because I am simply printing.

Answer 1

This is a precedence issue. *sptr[i] parses as *(sptr[i]) , but you need (*sptr)[i] .

It crashes because sptr[i] means *(sptr + i) , ie at i = 0 you're simply dereferencing sptr (giving you sentence ), then dereferencing that, giving you *sentence (ie the first character of sentence ). But at i = 1 you're doing sptr[1] (ie *(sptr + 1) ), which tries to access memory next to the sentence variable, then treats it as another pointer to dereference. Even if you're lucky and there happens to be a valid pointer next to sentence in memory, successive iterations will just attempt to dereference more and more garbage values as pointers.

Of course, you could avoid the whole problem by having printword take a char * and simply passing sentence to it.

Answer 2

Typical new C programmers mistake, thinking that *str[i] is the same as str[][i] it is not it is more like str[i][0] .

So, why is this happening? When you look at precedence [] is at the top. Take a look at this

The other way to look at it is:

arr[i][j]

is the like:

*(*(arr+i) + j)

thus *str[i] is like:

**(arr + i)

what you want is:

*(*(arr) + i)

which is like:

*(arr[i])

I hope this helps!!!!