sizeof 运算符的问题 [英] problem with sizeof operator

问题描述

因为我想在函数中动态查找数组大小，所以我使用了 sizeof 运算符.但我得到了一些意想不到的结果.这是一个演示程序，向您展示我想做什么.

As i want to find array size dynamically in function, i used sizeof operator. But i got some unexpected result. here is one demo program to show you, what i want to do.

//------------------------------------------------------------------------------------------
#include <iostream>

void getSize(int *S1){

    int S_size = sizeof S1/sizeof(int);
    std::cout<<"array size(in function):"<<S_size<<std::endl;
}

int main(){

    int S[]={1,2,3,2,5,6,25,1,6,21,121,36,1,31,1,31,1,661,6};
    getSize(S);
    std::cout<<"array size:"<<sizeof S/sizeof(int)<<std::endl;
    return 0;
}
//------------------------------------------------------------------------------------------

编译命令:g++ demo1.cc -o demo1 {fedora 12}

compilation command : g++ demo1.cc -o demo1 {fedora 12}

输出:

array size(in function):2
array size:19

<小时>

请解释，为什么会这样.可以做些什么来解决这个问题.

---

please explain ,why this is happening. what can be done to solve this problem.

推荐答案

void getSize(int *S1)

当你向这个函数传递一个数组时，它会衰减到pointer类型，所以sizeof操作符将返回指针的大小.

When you pass an array to this function, it decays to pointer type, so sizeof operator will return the size of pointer.

但是，您将函数定义为，

However, you define your function as,

template<int N>
void getSize(int (&S1)[N])
{
   //N is the size of array
   int S_size1 = N;
   int S_size2 = sizeof(S1)/sizeof(int); //would be equal to N!!
   std::cout<<"array size(in function):"<<S_size1<<std::endl;
   std::cout<<"array size(in function):"<<S_size2<<std::endl;
}

int S[]={1,2,3,2,5,6,25,1,6,21,121,36,1,31,1,31,1,661,6};
getSize(S); //same as before

那么你就可以在函数里得到数组的大小了！

then you can have the size of array, in the function!

在这里亲自观看演示:http://www.ideone.com/iGXNU

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