C 指针和数组:[警告] 赋值使指针来自整数而不进行强制转换 [英] C pointers and arrays: [Warning] assignment makes pointer from integer without a cast

问题描述

我在使用 C 中的指针和数组时遇到了一些问题.代码如下:

#includeint *ap;int a[5]={41,42,43,44,45};整数 x;int main(){ap = a[4];x = *ap;printf("%d",x);返回0；}

当我编译并运行代码时，我收到此警告:

<块引用>

[警告] 赋值使指针从整数而不进行强制转换[默认启用]

对于第 9 行(ap = a[4];)，终端崩溃.如果我将第 9 行更改为不包含位置 (ap = a;)，则不会收到任何警告并且可以正常工作.为什么会这样?我觉得答案很明显，但我就是看不到.

解决方案

在这种情况下 a[4] 是数组 a<中的 5 整数/code>，ap 是一个指向整数的指针，因此您将一个整数分配给一个指针，这是警告.
所以 ap 现在持有 45 并且当您尝试取消引用它时(通过执行 *ap)，您正在尝试访问地址 45 处的内存，这是一个无效地址，因此您的程序会崩溃.>

你应该做 ap = &(a[4]); 或 ap = a + 4;

在 c 中，数组名称衰减为指针，因此 a 指向数组的第一个元素.
这样，a 等价于 &(a[0]).

I'm having some trouble with pointers and arrays in C. Here's the code:

#include<stdio.h>


int *ap;
int a[5]={41,42,43,44,45};
int x;

int main()
{
    ap = a[4];
    x = *ap;
    printf("%d",x);
    return 0;
}

When I compile and run the code I get this warning:

[Warning] assignment makes pointer from integer without a cast [enabled by default]

For line number 9 (ap = a[4];) and the terminal crashes. If I change line 9 to not include a position (ap = a;) I don't get any warnings and it works. Why is this happening? I feel like the answer is obvious but I just can't see it.

解决方案

In this case a[4] is the 5th integer in the array a, ap is a pointer to integer, so you are assigning an integer to a pointer and that's the warning.
So ap now holds 45 and when you try to de-reference it (by doing *ap) you are trying to access a memory at address 45, which is an invalid address, so your program crashes.

You should do ap = &(a[4]); or ap = a + 4;

In c array names decays to pointer, so a points to the 1st element of the array.
In this way, a is equivalent to &(a[0]).

这篇关于C 指针和数组:[警告] 赋值使指针来自整数而不进行强制转换的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！