cpp空数组声明 [英] cpp empty array declaration
问题描述
您好,我有以下测试代码,我对cpp感到困惑。
Hello I have the following test code and I am confused about cpp.
-
如果您在library.h中声明数组有一个空的元素子句。编译器会选择什么?
If you declare in library.h an array with an empty element clause .. what will the compiler pick? It does also not complain, I use Cygwin.
在library.cpp中,我为两个元素赋值,是编译器假设一个数组有一个元素,I写入数组范围之外的第二个元素?
In library.cpp I assign values to two elements, is the compiler assuming an array with one element and I write the second element outside the scope of the array?
library.h
library.h
#ifndef LIBRARY_H
#define LIBRARY_H
class library {
public:
void print();
char a[];
};
#endif
library.cpp
library.cpp
#include <stdio.h>
#include "library.h"
void library::print() {
a[0] = 'a';
printf("1. element: %d\n", a[0]);
a[1] = 'b';
printf("2. element: %d\n", a[1]);
}
client.cpp
client.cpp
#include <stdio.h>
#include "library.h"
void execute();
library l;
int main() {
l = library();
l.print();
return 0;
}
Makefile
Makefile
OPTIONS=-Wall
all: main
run: main
./main.exe
main: client.o library.o
g++ $(OPTIONS) -o main $^
library.o: library.cpp library.h
g++ $(OPTIONS) -c $<
.cpp.o:
g++ $(OPTIONS) -c $<
clean:
rm -r *.o
推荐答案
- 没有C / C ++语言,因此您的Q不能同时标记两者。
- ,您的程序只能是C ++而不是C.
public:
void print();
char a[];
这段代码在C ++中是非法的。 C ++中的数组大小需要为正编译时常数。解决方案是替换它:
This code is simply illegal in C++. Array size in C++ needs to be positive compile time constant. Solution is to replace it by:
public:
void print();
std::string a;
请注意,声明
Note that the declaration,
char a[];
在c99中有效,称为不完整的阵列类型 C标准保证 a
可以存储 char
类型的至少一个元素。这在C ++中无效。 C ++标准不允许这些。只是因为两者都是不同的语言。
is valid in c99 and it is known as Incomplete array type, the C standard guarantees that a
can store atleast one element of the type char
. This is not valid in C++. C++ standard does not allow these. Simply because both are different languages.
这篇关于cpp空数组声明的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!