C语言复习 -- printf报错 Segmentation fault

刚才printf的时候，报错“Segmentation fault”。在网上有个例子写的比较详细，自己测试如下：

=== 代码：

#include <stdio.h>
#include <string.h> 
int main(void) {
    char buf[100];
    char *p = buf;   
    strcpy(p, "Test string");
    printf("%s\n", *p); 
   
    return 0;
}

=== 结果：

[oracle@aa tmp]$ ./a.out
Segmentation fault
[oracle@aa tmp]$

 

=== 分析原因：

When you write

printf("%s\n", *p); 

the *p will be the value at p[0] which is a character. The printf however is looking for an array of chars, thus causing it to segfault. Remember that in C, strings are just arrays of characters, and arrays are effectively pointers to the first element, this is why you don't need to dereference.

To fix this remove the * to get:

printf("%s\n", p); 

 

 

=== 修改重测：

[oracle@AA tmp]$ vi a.c
[oracle@AA tmp]$ gcc a.c
[oracle@AA tmp]$ ./a.out
Test string
[oracle@AA tmp]$

原址如下：http://stackoverflow.com/questions/2976204/c-segmentation-fault-while-using-printf