为什么sprintf和snprintf的行为是不同的，当我们使用相同的源和目的？ [英] why Behavior of sprintf and snprintf is different when we use same source and destination?

问题描述

我有一个简单的code，它使用的sprintf

I have a simple code which uses sprintf

    #include <stdio.h>

    int main()
    {

            char str_src [1024]={"Hello"};
            sprintf(str_src,"%s%s",str_src,"hiiiiiiiiiii");
            printf("result = %s",str_src);  

    }

我编译时得到正确的结果：

When i compile i get correct result :

结果= Hellohiiiiiiiiiii

result = Hellohiiiiiiiiiii

但由于sprintf的是不安全，我决定改变这snprintf的。我认为这将是非常简单的。我改变的sprintf到的snprintf像下面

But since sprintf is unsecure, i decided to change this to snprintf. I thought it would be really simple. I changed sprintf to snprintf like below

snprintf(str_src,1024,"%s%s",str_src,"hiiiiiiiiiii");

现在如果我编译并运行code，我得到不同的结果。

Now If i compile and run the code, i get different result

结果= hiiiiiiiiiii

result = hiiiiiiiiiii

我面对这个问题，如果我使用str_src作为第四个参数（作为值％S）。它令人惊奇为什么snprintf的行为比sprintf的有什么不同？

I face this problem if i use str_src as 4th parameter (as a value to %s). Its suprising why the behavior of snprintf is different than sprintf?

推荐答案

这是未定义行为的使用相同的缓冲既作为目的地和来源。

It's undefined behavior to use the same buffer both as destination and source.

从C11规范（7.21.6.6/2）：

From the C11 specification (7.21.6.6/2):

如果复制操作是重叠的对象之间，行为是不确定的。

If copying takes place between objects that overlap, the behavior is undefined.

同样是说，的snprintf （7.21.6.5/2），并且也对的va_list 变体好。

The same is said for snprintf (7.21.6.5/2), and also on the va_list variants as well.

不幸的是这一切都共同运行code，但它不能真正依靠工作。

Unfortunately it's all to common in running code, but it can't really be relied on to work.

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