无法理解 C++ 指针语法 [英] Trouble understanding C++ pointer syntax

问题描述

我无法理解我在面试中遇到的这段代码声明.

I'm not able to understand this statement of code that I came across during my interview.

int(*(*ptr[3])(char*))[2];

我曾尝试查看 IDE，但我所拥有的只是它是一个数据类型数组

I've tried looking at an IDE but all I have is that it is an array of data type

int (*(*[3])(char *)) 

我无法理解这一点.

推荐答案

也许你可以一次分解一个来更好地理解语法.首先从一个没有数组符号的简单定义开始

May be you could just break it down one at a time to understand the syntax better. First start up with a simple definition without the array notation

int(*(*ptr)(char*));

所以ptr 是一个函数指针，它接受一个char 指针作为参数并返回一个指向int 的指针.现在将其扩展为数组符号

So ptr is a function pointer that takes a char pointer as an argument and returns a pointer to an int. Now extending it to the array notation

int(*(*ptr[3])(char*))[2];

这意味着你有一个函数指针数组，每个函数指针都接受一个 char 指针参数并返回一个指向两个整数数组的指针.

which means you have an array of function pointers, each of which will take a char pointer argument and return a pointer to an array of two integers.

如果您使用您定义的这些指针进行函数调用，您可以看到此功能.请注意，以下函数仅用于演示目的，不传达任何逻辑目的

You can see this working if you have a make a function call using these pointers you define. Note that, the below functions are for demonstrative purposes only and do not convey any logical purpose

#include <iostream>

static int arr[2] = { 2, 2 };

// initialize  'bar' as a function that accepts char* and returns
// int(*)[2]
int (*bar(char * str))[2] {
    return &arr;
}

int main() {
    // pointer definition, not initialized yet
    int(*(*foo[3])(char*))[2];
    char ch = 'f';
    // as long as the signatures for the function pointer and 
    // bar matches, the assignment below shouldn't be a problem
    foo[0] = bar;
    // invoking the function by de-referencing the pointer at foo[0]
    // Use 'auto' for C++11 or declare ptr as int (*ptr)[2] 
    auto *ptr = (*foo[0])(&ch);
    return 0;
}

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