c - 编译时段错误的可能原因

Question

问题:

我的程序正在运行，但未成功。它返回段错误。可能是哪里出了问题或者如何解决这个问题？

代码:

#include<stdio.h>
#include<stdlib.h>
int main ()
{
  int *arr, i, j, n;
  printf ("Input the Size");
  scanf ("%d", n);
  arr=(int*)malloc(n*sizeof(int));
  for (i = 0; i < n, i++;)
  {
    *(arr + i) = i;
  }
  for (i = 0; i < n, i++;)
  {
    printf ("%d\n", *(arr + i));
  }
return 0;
}

Answer 1

更改此:

scanf ("%d", n);

对此:

scanf ("%d", &n);

自 scanf( const char * format, ...) 需要一个指针作为其参数。

<小时/>

此外更改此:

for (i = 0; i < n, i++;)

对此:

for (i = 0; i < n; i++)

因为通常的 for 循环语法是:

for ( init; condition; increment )

第二个 for 循环相同:for (i = 0; i < n, i++;) ，应该是 for (i = 0; i < n; i++)

<小时/>

您可以通过传递编译启用标志(例如Wall)来允许编译器警告您这一点。对于 GCC(像这样编译: gcc prog.c -Wall )。然后您会收到如下警告:

prog.c: In function 'main':
prog.c:7:12: warning: format '%d' expects argument of type 'int *', but argument 2 has type 'int' [-Wformat=]
    7 |   scanf ("%d", n);
      |           ~^   ~
      |            |   |
      |            |   int
      |            int *
prog.c:9:20: warning: left-hand operand of comma expression has no effect [-Wunused-value]
    9 |   for (i = 0; i < n, i++;)
      |                    ^
prog.c:13:20: warning: left-hand operand of comma expression has no effect [-Wunused-value]
   13 |   for (i = 0; i < n, i++;)
      |                    ^
prog.c:5:16: warning: unused variable 'j' [-Wunused-variable]
    5 |   int *arr, i, j, n;
      |                ^
prog.c:7:3: warning: 'n' is used uninitialized in this function [-Wuninitialized]
    7 |   scanf ("%d", n);
      |   ^~~~~~~~~~~~~~~