分割故障 - Ç [英] Segmentation Fault - C

问题描述

为什么以下code返回段故障？当我注释掉线7​​条，赛格故障消失。

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)] = '\0';
      puts(dyn_s);
      return 0;
    }

干杯！

推荐答案

取值是一个未初始化的指针;您在内存中写入随机位置。这将调用的未定义行为

s is an uninitialized pointer; you are writing to a random location in memory. This will invoke undefined behaviour.

您需要分配一些内存为取值。此外，从来不使用获得 ;有没有办法为prevent它溢出你分配的内存。使用与fgets 代替。

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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