C指针和数组/'的sizeof“运算符 [英] C pointers and arrays/ 'sizeof' operator

问题描述

可能重复：
  对于字符指针和阵列堆栈指针的区别

要说明我的问题：

int main(void){
    int myary[20];
    int *myaryPtr;
    myaryPtr = myary;

    sizeof(myary); // Will it return 80? Correct?
    sizeof(myaryPtr); // Will it return 4? Correct?
    return 0;
}

首先，是我的假设是正确的？

First off, is my assumption correct?

再假设我的假设是正确的，什么是详细的解释？我明白我的20个元素的数组是80个字节，而不是名称 myary 只是一个指向数组的第一个元素？所以，不应该也为4？

And then assuming my assumption is correct, what is the detailed explanation? I understand that my 20 element array is 80 bytes, but isn't the name myary merely a pointer to the first element of the array? So shouldn't it also be 4?

推荐答案

是的，你的假设是正确的，假设一个 INT 和指针都是4个字节长在你的机器。

Yes, your assumption is correct, assuming an int and a pointer are both 4 bytes long on your machine.

和没有，数组不是指针。数组名有时会衰变成在某些情况下的指针，但它们不是一回事。有专门为混乱的公共点的comp.lang.c常见问题的整节。

And no, arrays aren't pointers. The array name sometimes decays into a pointer in certain contexts, but they aren't the same thing. There is a whole section of the comp.lang.c FAQ dedicated to this common point of confusion.

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