If you assign a string to the character variable, it may cause a warning or error (in some of the compilers) or segmentation fault error occurs.

如果將字符串分配給字符變量，則可能會導致警告或錯誤(在某些編譯器中)或發生分段錯誤。

Consider the code:

考慮一下代碼：

#include <stdio.h>
int main(void) {
char name="Amit shukla";
printf("%s",name);
return 0;
}

Output

輸出量

Segmentation fault

How to fix?

怎么修？

Declare character array instead of char variable to assign string

聲明字符數組而不是char變量來分配字符串

 char name[]="Amit shukla";

翻譯自: https://www.includehelp.com/c-programs/assign-string-to-the-char-variable-in-c.aspx