How to print a char array in C through printf? [closed]

Question

This results in segmentation fault. What needs to be corrected?

int main(void) {     char a_static = {'q', 'w', 'e', 'r'};     char b_static = {'a', 's', 'd', 'f'};      printf("\n value of a_static: %s", a_static);     printf("\n value of b_static: %s\n", b_static); } 

Answer 1

The code posted is incorrect: a_static and b_static should be defined as arrays.

There are two ways to correct the code:

• you can add null terminators to make these arrays proper C strings:

  #include <stdio.h>  int main(void) {     char a_static[] = { 'q', 'w', 'e', 'r', '\0' };     char b_static[] = { 'a', 's', 'd', 'f', '\0' };      printf("value of a_static: %s\n", a_static);     printf("value of b_static: %s\n", b_static);     return 0; } 

• Alternately, printf can print the contents of an array that is not null terminated using the precision field:

  #include <stdio.h>  int main(void) {     char a_static[] = { 'q', 'w', 'e', 'r' };     char b_static[] = { 'a', 's', 'd', 'f' };      printf("value of a_static: %.4s\n", a_static);     printf("value of b_static: %.*s\n", (int)sizeof(b_static), b_static);     return 0; } 

  The precision given after the . specifies the maximum number of characters to output from the string. It can be given as a decimal number or as * and provided as an int argument before the char pointer.

Answer 2

This results in segmentation fault. ? because of the below statement

char a_static = {'q', 'w', 'e', 'r'}; 

a_static should be char array to hold multiple characters. make it like

 char a_static[] = {'q', 'w', 'e', 'r','\0'}; /* add null terminator at end of array */ 

Similarly for b_static

char b_static[] = {'a', 's', 'd', 'f','\0'};