本文介绍了分段错误 - C的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
为什么以下代码返回分段错误?当我注释掉第 7 行时，段错误消失了.
Why does the following code return with a segmentation fault? When I comment out line 7, the seg fault disappears.
int main(void){
char *s;
int ln;
puts("Enter String");
// scanf("%s", s);
gets(s);
ln = strlen(s); // remove this line to end seg fault
char *dyn_s = (char*) malloc (strlen(s)+1); //strlen(s) is used here as well but doesn't change outcome
dyn_s = s;
dyn_s[strlen(s)] = ' ';
puts(dyn_s);
return 0;
}
干杯！
推荐答案
s是一个未初始化的指针；您正在写入内存中的随机位置.这将调用未定义的行为.
s is an uninitialized pointer; you are writing to a random location in memory. This will invoke undefined behaviour.
您需要为 s 分配一些内存.另外，永远不要使用gets；没有办法防止它溢出您分配的内存.请改用 fgets.
You need to allocate some memory for s. Also, never use gets; there is no way to prevent it overflowing the memory you allocate. Use fgets instead.
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