用C动态数组没有malloc的？ [英] Dynamic arrays in C without malloc?

问题描述

我一直在想，我怎么能逃脱这样的：

I've always wondered how I could get away with this:

int main(int argc, char **argv) {
    printf("%p %s %d\n", &argv[1], argv[1], strlen(argv[1]));
    char copy[strlen(argv[1]) + 1];
    strcpy(copy, argv[1]);
    printf("%p %s %d\n", &copy, copy, strlen(copy));
    return 0;
}

该字符数组复制被分配无论如何，程序运行正常，打印出的原件和复印件。而Valgrind的不抱怨什么。

The char array copy gets allocated anyway and the program runs fine, printing out the original and the copy. And Valgrind doesn’t complain about anything.

我想的动态的阵列是不可能在C时不malloc的。难道我错了吗？

I thought dynamic arrays weren’t possible in C without malloc. Was I wrong?

推荐答案

这是C99的功能，并且可以在的以前版本的编译器。

This is a C99 feature and could be implemented on prior versions by the compiler.

变长自动数组被允许在ISO C99，并作为
  扩展GCC接受他们C90模式和C ++。这些阵列
  宣布像任何其他自动数组，但具有的长度是
  不是一个常数前pression。存储在点分配
  声明和解除当护具级退出。

Variable-length automatic arrays are allowed in ISO C99, and as an extension GCC accepts them in C90 mode and in C++. These arrays are declared like any other automatic arrays, but with a length that is not a constant expression. The storage is allocated at the point of declaration and deallocated when the brace-level is exited.

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