printf 在特定上下文中不起作用，为什么？

Question

I needed to test something and programmed this little bit of code(shown below). I can't understand why the first print works and the second doesn't. The output of this program is just

    this prints

but it should be

    this prints
    this doesn't print i: 1

Here's the code:

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

int cmp(char *str) {
    char *z[1];
    strcpy(*z, "z");
    int a;

    a = strcmp(str, *z);

    return a;
}

int main() {
    int i;
    char *name[1];
    printf("this prints\n");

    strcpy(*name, "y");
    i = cmp(*name);
    printf("this doesn't print i:%d", i);
    return 0;
}

Answer 1

char *z[1]; // this is an array of pointer
strcpy(*z, "z"); // you have to allocate at least 2 bytes for *z

// and
char *name[1];
strcpy(*name, "y"); // you have to allocate at least 2 bytes for *name also

You did not allocate for the pointer in array z and name .

Your cmp function looks weird. If you want to compare string with "z" , you can just do:

int cmp(char *str){
   return strcmp(str, "z");
}

You do not need to use char *name[1] , just char *name = malloc(SIZE+1); or char name[SIZE+1] ( SIZE is length of string that you want to compare) is enough.

Answer 2

char *z[1]; and char *name[]

1. Both name and z are not arrays of char . They are both an array of one pointer to char .

2. Both pointers name[1] and z[1] point to no valid memory, thus dereferencing it and attempting to store a string to that undefined memory by using strcpy(*name, "y"); and strcpy(*z, "z"); and invokes undefined behavior . - What you need is an array of char for both, name and z .

3. For storing a string you need one element to store the string-terminating null character.

Use char z[2] and name[2] , if you only want to have strings contained of one single character.

---

BTW, Your code is a little bit complicate handled. You do not to use strings if you want to store and compare only single characters. It could be simplified as:

#include <stdio.h>

int main (void) {

    char name = 'y';
    printf("In name is the character: '%c'\n", name);

    printf("Is the character 'z' in 'name'? %d", name == 'z');
    return 0;
}

Output:

In name is the character: 'y'
Is the character 'z' in 'name'? 0

Where 0 means false and 1 signifies true .

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Side Note:

You didn´t need to #include <stdlib.h> at any point of time.