为什么这个segc会出错？

我不明白为什么这个小小的C程序会出错：

#include <stdio.h>
#include <stdlib.h>

int main(int argc, char *argv[]){

    int in = atoi(argv[1]);
    printf("Input %d\n",in);

    int *n = (int *)malloc(in);
    int j;

    for (j=0;j<in;j++)
        n[j] = j;

    printf("Sanity check...\n");

    char *c = (char *)malloc(1024*1024*20);
    int i;
    for (i=0; i<20*1024*1024;i++)
        c[i] = i;

    printf("No segfault. Yay!\n");

    return 0;
}

#包括
#包括
int main（int argc，char*argv[]）{
int in=atoi（argv[1]）；
printf（“输入%d\n”，in）；
int*n=（int*）malloc（in）；
int j；
对于（j=0；j
您正在以

字节分配

，而不是以

整数分配

。请尝试：

malloc(sizeof(int) * in);

您的第二次分配有效，因为
sizeof（char）
是
1
此语句

int *n = (int *)malloc(in);

这是错误的

我想你是说

int *n = (int *)malloc(in * sizeof( int ) );

给定发布的代码，
让我们来分析一下它出现故障的原因。
我的评论中夹杂着代码
#包括
#包括
//未使用argc，因此编译器将发出警告
int main（int argc，char*argv[]）
{
//未检查argc的值以确保参数存在
//因此，当不存在任何参数时，不会输出“用法”消息
//或者存在太多的参数
//注意：'in'是（IMO）局部变量的糟糕名称
int in=atoi（argv[1]）；
//如果“in”大于0，则缺少确保值的检查
//当值也是时，处理错误时，不鼓励使用
malloc
。@MichałSzydłowski为什么不鼓励使用malloc？如果没有有效的解释，你不能轻视有效的答案。我不轻视答案-这个答案纠正了错误，因此不再存在分段错误。它会响应关于这个问题。我的补充意见是，您编写的代码应该尽可能独立于类型-这意味着避免不必要的强制转换。在这里使用
malloc
的强制转换无论如何都是隐式的，因此您可以让编译器毫无问题地执行它是的，谢谢你，我已经做了更改，它按预期工作。谢谢！我编写了这个小程序只是为了测试。它代表了我实际编写的流程，以便找出故障发生的位置和原因，而不必处理“生产”中的所有其他复杂性代码。您提到的格式化内容（例如检查argc值和处理错误）是在我正在编写的实际应用程序中处理的——我愚蠢地忽略了int数组的malloc，并从中了解到了不进行malloc类型转换。因此，感谢您的输入！
int *n = (int *)malloc(in);

int *n = (int *)malloc(in * sizeof( int ) );

given the posted code,
lets just run down the reasons it seg faults.

my comments are interspersed with the code

#include <stdio.h>
#include <stdlib.h>

// argc is not used, so the compiler will raise a warning
int main(int argc, char *argv[])
{

    // the value of argc was not checked to assure a parameter is present
    // so a 'usage' message was not output when no parameter is present
    // or too many parameters are present
    // note: 'in' is (IMO) a poor name for a local variable
    int in = atoi(argv[1]);
    // missing check to assure the value if 'in' is greater than 0
    // and handling the error when the value is <= 0

    printf("Input %d\n",in);

    // in C, the returned value from calls to malloc(), and family,
    //       should not be cast
    // the returned value is (per the definition of 'n')
    // to be a pointer to an array of int's
    // which should be written as:
    // int *n = malloc( in * sizeof int );
    // the returned value from malloc(), and family of functions,
    //     needs to always be checked (!= NULL)
    //     to assure the operation was successful
    int *n = (int *)malloc(in);
    int j;

    // because the malloc call did not ask for a memory allocation
    // of 'in' int's, so this code block corrupts the heap
    // by writing past the end of the allocated memory
    for (j=0;j<in;j++)
        n[j] = j;

    printf("Sanity check...\n");

    // in C, the returned value from calls to malloc(), and family of functions,
    //       should not be cast
    // the returned value needs to be checked (!= NULL) to assure the operation
    //       was successful
    // the calculated value: 1024*1024*20 should be calculated by the
    //       compiler, not repeatedly at run time.
    //       so insert 'const int twentyMillion = 1024*1024*20;'
    //       at beginning of program
    char *c = (char *)malloc(1024*1024*20);
    int i;

    // to avoid repeatedly calculating the 20million value, use a const (as above)
    // so this line would be: for ( i=0; i<twentyMillion; i++ )
    for (i=0; i<20*1024*1024;i++)
        // the value of 'i' will massively exceed the value that can be kept in a char
        //  I.E. 255 so this loses data
        //  suggest replacing following line with: j = i%256;  c[i] = j;
        // to make it obvious what is going on
        c[i] = i;

    printf("No segfault. Yay!\n");

    return 0;
} // end function: main