只是为了测试我创建了以下代码：
#include<stdio.h> int main(){ char *p = "Hello world"; *(p+1) = 'l'; printf("%s",p); return 0; }
但当我在ubuntu 10.04下运行我的“gcc”编译器时,我得到了：
Segmentation fault
所以任何人都可以解释为什么会这样.
#include<stdio.h> #include<stdlib.h> int main(){ char *p = malloc(sizeof(char)*100); p = "Hello world"; *(p+1) = 'l'; printf("%s",p); free(p); return 0; }
这也会导致分段错误
提前致谢
解决方法
char *p = "Hello world";
*(p+1) = 'l';
修饰字符串文字的内容(即代码中的“Hello World”)是未定义的行为.
ISO C99(第6.4.5 / 6节)
It is unspecified whether these arrays are distinct provided their elements have the appropriate values. If the program attempts to modify such an array,the behavior is undefined.
尝试使用字符数组.
char p[] = "Hello World"; p[1] = 'l';
编辑
#include<stdio.h> #include<stdlib.h> int main() { char *p = malloc(sizeof(char)*100); p = "Hello world"; // p now points to the string literal,access to the dynamically allocated memory is lost. *(p+1) = 'l'; // UB as said before edits printf("%s",p); free(p); //disaster return 0; }
也会调用未定义的行为,因为您正在尝试释放尚未使用malloc分配的内存部分(使用free)
