打印指向整数的指针会导致分段错误.为什么? [英] Printing pointer to integer causes segmentation fault. Why?

问题描述

#include<stdio.h>
main()
{
    int *num2=20;
    printf("


%d",num2);
}

当我按原样运行它时，它会打印 20.如果我使用 *num2 它会导致分段错误.为什么?

When I run it as it is, it prints 20. If I use *num2 it causes segmentation fault. Why?

推荐答案

为什么打印 num 没问题，但 *num 导致分段错误?

当你说，int *num2=20;.相当于

Why printing num was okay but *num resulted in segmentation fault?

When you say, int *num2=20;. It is equivalent to

int *num; /* Type:(int*) num is a pointer to an integer and hence holds address of an integer object. */
num = 20; /* Type of the value assigned to num is (int).*/

您将 int 分配给 int * .你应该收到一个

You are assigning int to int * . You should have received a

warning: initialization makes pointer from integer without a cast[enabled by defaut]

num 指向地址20".即使那是一个有效地址，你也不知道.

num points to the address '20'. You have no idea even if that is a valid address.

因此，当您仅打印 num 时，您没有任何问题.但是取消引用，地址(20)是无效的内存读取，导致未定义的行为，并导致分段错误.

Hence when you printed just num you had no problems. But dereferencing, the address(20) is invalid memory read , causes undefined behaviour and that resulted in segmentation fault.

num 应该包含一个有效的地址.不要冒险自己分配一些随机地址.

num should hold a valid address. Do not take this risk by assigning some random address yourself.

通过创建 int 类型的对象并分配 num 来保存它的地址或为此分配内存来避免这种风险.喜欢，

Avoid that risk by either creating an object of type int and assign num to hold the address of it or just allocate memory for that. Like,

i):

int var = 20;    /* var is a integer object holding the value 20 */

& 一元运算符可帮助您获取操作数的地址.使用 & 和 var 来获取它的地址，因为 var 是一个整数，所以将它存储在一个整数指针中.在你的情况下，num.

& unary operator helps you get the address of the operand. Use & with var to get its address and since var is an integer, store it in an integer pointer. In your case, num.

int *num = &var; /* num is a pointer to var. Holds the address of var. */

ii):

int *num = malloc(sizeof *num);
*num = 20; 

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