为什么函数不知道数组大小？ [英] Why function does not know the array size?

问题描述

如果我写

int main()
{
    int a[100] = {1,2,3,4,};
    cout<<sizeof(a)/sizeof(a[0])<<endl; //a is a pointer to the first elem of array, 
                                        //isn't it
    return 0;
}

我得到400！

如果我写

void func（int * a）;

void func(int *a);

int main()
{
    int a[100] = {1,2,3,4,};
    func(a);
    return 0;
}

void func(int *a)
{
     cout<<sizeof(a)/sizeof(a[0])<<endl; //a is a pointer to the first elem of array
}

那么，为什么函数不知道数组大小？

So why function does not know the array size?

推荐答案

这不工作，因为sizeof是在编译时计算的。该函数没有关于它的参数大小的信息（它只知道它指向一个内存地址）。

This isn't working because sizeof is calculated at compile-time. The function has no information about the size of its parameter (it only knows that it points to a memory address).

考虑使用STL向量，或者传入数组尺寸作为函数的参数。

Consider using an STL vector instead, or passing in array sizes as parameters to functions.

这是由Marcel Guzman在 http://stackoverflow.com/questions/720077/calculating-size-of-an-array ！

This was answered by Marcel Guzman in http://stackoverflow.com/questions/720077/calculating-size-of-an-array!

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