查找保留内存的大小在C语言的字符数组 [英] Find the size of reserved memory for a character array in C

问题描述

我想学C，并作为一个开始，我掀起写我自己的实践中的strcpy。我们知道，原来的strcpy容易使得安全问题，所以我给自己写了一个安全的strcpy任务。

I'm trying to learn C and as a start, i set off writing a strcpy for my own practice. As we know, the original strcpy easily allows for security problems so I gave myself the task to write a "safe" strcpy.

我选择的路径是请检查是否源字符串（字符数组）的目标内存实际相符。正如我的理解是，在C字符串无非就是一个指针更上一个字符数组，为0x00终止。

The path I've chosen is to check wether the source string (character array) actually fits in the destination memory. As I've understood it, a string in C is nothing more than a pointer to a character array, 0x00 terminated.

所以我的挑战是如何找到编译器实际上多少内存预留给目标字符串？

So my challenge is how to find how much memory the compiler actually reserved for the destination string?

我试过：

sizeof(dest)

但是，这并不工作，因为它会返回（因为我后来发现）dest的大小，这实际上是一个指针，我的64位计算机上，总是会返回8。

but that doesn't work, since it will return (as I later found out) the size of dest which is actually a pointer and on my 64 bit machine, will always return 8.

我也试过：

strlen(dest)

但是，这并不工作，可能是因为它会直接返回长度遇到的第一个为0x0到，这并不一定反映实际的内存保留。

but that doesn't work either because it will just return the length until the first 0x0 is encountered, which doesn't necessarily reflect the actual memory reserved.

因此​​，这所有款项到以下问题：如何找到我们的编译器多少内存留给我的目的地串???

So this all sums up to the following question: How to find our how much memory the compiler reserved for my destination "string"???

例如：

char s[80] = "";
int i = someFunction(s); // should return 80

什么是someFunction？

What is "someFunction"?

在此先感谢！

推荐答案

您可以用sizeof的检查在编译时：

You can use sizeof to check at compile time:

char s[80] = "";
int i = sizeof s ; // should return 80

请注意，如果s是一个指针失败：

Note that this fails if s is a pointer:

char *s = "";
int j = sizeof s;  /* probably 4 or 8. */

数组不是指针。若要跟踪分配的指针大小，程序则只是必须保持它的轨道。此外，您不能传递一个数组给一个函数。当您使用数组作为参数传递给一个函数，编译器转换，为一个指向第一个元素，所以如果你想要的大小是avaliable所调用的函数，它必须作为参数传递。例如：

Arrays are not pointers. To keep track of the size allocated for a pointer, the program simply must keep track of it. Also, you cannot pass an array to a function. When you use an array as an argument to a function, the compiler converts that to a pointer to the first element, so if you want the size to be avaliable to the called function, it must be passed as a parameter. For example:

char s[ SIZ ] = "";
foo( s, sizeof s );

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