差异:& ary1D [0]和& ary1D [英] Difference: &ary1D[0] and &ary1D

问题描述

我的主函数中有一个双数组:

I have in my main-Function an double Array:

double ary1D[4] = {1.1, 2.2, 3.3, 4.4};

并进行函数调用:

print1D_A(&ary1D[0],Num);

我的功能:

void print1D_A(double *ary1D, int Num);

所以我的问题是，& ary1D [0] 和& ary1D 有什么区别.如果我使用& ary1D 调用函数，编译器会给我一个错误.但是参数& ary1D 是我数组的第一个地址，就像& ary1D [0] 一样.

so my question is, Whats is the difference between &ary1D[0] and &ary1D. My compiler give me an error, if I call the function with &ary1D. But the argument &ary1D is the first Adress of my Array, just like &ary1D[0].

我可以更改函数的参数列表吗，可以用

Can I change the argumentlist of my function, that I can call it with

print1D_A(&ary1D,Num);

推荐答案

表达式&ary1D[0]是指向ary1D的第一个元素的指针.其类型为double*.这也是ary1D 衰变的原因，所以您可以这样做

The expression &ary1D[0] is a pointer to the first element of ary1D. Its type is double*. This is also what plain ary1D decays to, so you could just do

print1D_A(ary1D,Num);

表达式&ary1D是指向 array 的指针.它的类型是double (*)[4].

The expression &ary1D is a pointer to the array. Its type is double (*)[4].

现在，两个指针都指向相同的位置，但是如您所见，它们的类型非常不同.如果一个函数需要一种类型而您又给了另一种类型，则可能会导致错误.

Now, both pointers are pointing to the same location, but as you can see their types are very different. That could lead to errors if a function is expecting one type and you give the other.

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