我们可以给静态数组的大小一个变量 [英] can we give size of static array a variable

问题描述

你好每一个我想问我，我已经阅读，我们可以声明动态数组只使用指针和使用malloc或newlike

hello every one i want to ask that i have read that we can declare dynamic array only by using pointer and using malloc or newlike

int * array = new int[strlen(argv[2])];

但我写了

int array[strlen(argv[2])];

它没有错误。

i已经读过静态数组只能通过给定常量数组大小来声明，但这里我给了一个可变大小的静态数组

i have read that static array can only be declared by giving constant array size but here i have given a variable size to static array

为什么是这样感谢

是安全使用或有机会，在任何后期阶段，它会使我的问题，我使用gcc linux

is it safe to use or is there chance that at any latter stages it will make problem i am using gcc linux

推荐答案

有什么称为可变长数组（VLA），它不是C ++的一部分， 是C99的一部分。许多编译器提供此功能作为扩展。

What you have is called a variable-length array (VLA), and it is not part of C++, although it is part of C99. Many compilers offer this feature as an extension.

即使是非常新的C ++ 11不包括VLA，因为整个概念不适合高级类型系统的C ++ 11（例如 decltype（array）？），C ++为运行时大小的数组提供了开箱库解决方案，更强大（如 std :: vector ）。

Even the very new C++11 doesn't include VLAs, as the entire concept doesn't fit well into the advanced type system of C++11 (e.g. what is decltype(array)?), and C++ offers out-of-the box library solutions for runtime-sized arrays that are much more powerful (like std::vector).

在GCC中，使用 std = c ++ 98 / c ++ 03 / c ++ 0x 和 -pedantic 将给予警告。

In GCC, compiling with -std=c++98/c++03/c++0x and -pedantic will give you a warning.

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