传递给非主函数的数组上基于范围的 for 循环

当我尝试在 gcc 4.8.2 中编译以下代码时，出现以下错误:

<块引用>

test.cc: 在函数‘void foo(int*)’中:test.cc:15:16: 错误:没有匹配的函数调用‘begin(int*&)’for (int i : bar) {^

以及来自模板库深处的其他一些人.

#include 使用命名空间标准；void foo(int*);int main() {int bar[3] = {1,2,3};for (int i : bar) {cout<<我＜＜结束;}富(酒吧)；}void foo(int* bar) {for (int i : bar) {cout<<我＜＜结束;}}

如果我重新定义 foo 以使用索引的 for 循环，那么代码会编译并按预期运行.此外，如果我将基于范围的输出循环移动到 main 中，我也会得到预期的行为.

如何将数组 bar 传递给 foo 使其能够在其上执行基于范围的 for 循环?

解决方案

随着数组衰减变成一个指针，你'丢失了一条重要信息:它的大小.

使用数组引用，您的基于范围的循环可以工作:

void foo(int (&bar)[3]);int main() {int bar[3] = {1,2,3};for (int i : bar) {cout<<我＜＜结束;}富(酒吧)；}void foo(int (&bar)[3]) {for (int i : bar) {cout<<我＜＜结束;}}

或者，以通用方式(即在函数签名中不指定数组大小)，

模板void foo(int (&bar)[array_size]) {for (int i : bar) {cout<<我＜＜结束;}}

试试

When I try to compile the following code in gcc 4.8.2, I get the following error:

test.cc: In function ‘void foo(int*)’:
test.cc:15:16: error: no matching function for call to ‘begin(int*&)’
   for (int i : bar) {
                ^

Along with a bunch of others from deeper in the template library.

#include <iostream>
using namespace std;

void foo(int*);

int main() {
  int bar[3] = {1,2,3};
  for (int i : bar) {
    cout << i << endl;
  }
  foo(bar);
}

void foo(int* bar) {
  for (int i : bar) {
    cout << i << endl;
  }
}

If I redefine foo to use an indexed for loop, then the code compiles and behaves as expected. Also, if I move the range-based output loop into main, I get the expected behaviour as well.

How do I pass the array bar to foo in such a way that it is capable of executing a range-based for-loop on it?

解决方案

With the array decaying into a pointer you're losing one important piece of information: its size.

With an array reference your range based loop works:

void foo(int (&bar)[3]);

int main() {
  int bar[3] = {1,2,3};
  for (int i : bar) {
    cout << i << endl;
  }
  foo(bar);
}

void foo(int (&bar)[3]) {
  for (int i : bar) {
    cout << i << endl;
  }
}

or, in a generic fashion (i.e. without specifying the array size in the function signature),

template <std::size_t array_size>
void foo(int (&bar)[array_size]) {
  for (int i : bar) {
    cout << i << endl;
  }
}

Try it out