snprintf错误。 sizeof的参数与目的地相同 [英] snprintf error. argument to sizeof is the same as destination

问题描述

当我构建时，gcc 4.8给我一个错误

  #include< string.h> 
 #include< stdio.h> 
 
静态内联void toto（char str [3]）
 {
 snprintf（str，sizeof（str），XX）; 
} 
 
 int main（）{
 char str [3]; 
 toto（str）; 
返回0; 
} 
  

以下是gcc错误

错误：'snprintf'调用中的'sizeof'参数与目标表达式相同;你的意思是提供一个明确的长度吗？

注意：我使用-Wall -Werror标志将警告转换为错误。
$ b

这里有类似的东西
在一个评论中，有人回答了这个问题

对于固定长度的缓冲区，我通常使用strncpy（dest，src，sizeof（dest））; dest [ sizeof（dest）-1] ='\0';这保证了NULL终止，并且比snprintf更简单，更不用说很多人使用snprintf（dest，sizeof（dest），src）;而是非常

但这是错误的：
gcc 4.8说

错误：'strncpy'调用中的'sizeof'参数与目标表达式相同;您的意思是提供明确的长度吗？[-Werror = sizeof-pointer-memaccess] strong>

在gcc 4.8文档中，他们正在讨论这个问题：
他们说：

-Wall的行为已经改变，现在包含新的警告标志-Wsizeof-pointer-memaccess。这可能会导致代码中出现新的警告，这些警告与以前版本的GCC一起编译干净。

例如，

  include string.h 
 
 struct A {}; 
 
 int main（void）
 {
 obj; 
 A * p1 =& obj; 
 A p2 [10]; 
 
 memset（p1，0，sizeof（p1））; //错误
 memset（p1，0，sizeof（* p1））; // ok，取消引用
 memset（p2，0，sizeof（p2））; //好的，数组
返回0; 
 
 
 
 
 
给出以下诊断：
 warning：argument to 'sizeof'in'void  memset（void *，int，size_t）'call与目标表达式相同;你的意思是解除引用吗？ [-Wsizeof-pointer-memaccess] 
 memset（p1，0，sizeof（p1））; //错误
 ^ 
尽管这些警告不会导致编译失败，但是经常将-Wall与-Werror结合使用，因此，新的警告将转化为新的错误。 
要解决这个问题，要么重写为使用memcpy，要么取消引用有问题的memset调用中的最后一个参数。* 
 
 
 
那么，在他们的例子中，代码是错误的，但在我的情况下，与snprintf / strncpy，我不明白为什么，我认为这是一个gcc错误的positif错误。 
 
 
 
感谢您的帮助
解决方案
传递给函数时指向第一个元素的指针。所以你有什么 
 
 
  static inline void toto（char str [3]）{..}  
 
 
 
不是一个数组，而是一个指针。
 
因此，gcc正确地警告。
 
 
 
是否在函数参数中指定大小并不重要： 
 
 
  static inline void toto（char str [3]）
  
和
  static inline void toto（char str []）
  
和
  static inline void toto（char * str）
  
都是相同的。
 
 
 
请阅读此处： 数组衰减？  
 
gcc 4.8 give me an error when I build 
#include <string.h>
#include <stdio.h>

static inline void toto(char str[3])
{
    snprintf(str, sizeof(str), "XX"); 
}

int main(){
    char str[3]; 
    toto(str);
    return 0;
}
Here is the gcc error

error: argument to ‘sizeof’ in ‘snprintf’ call is the same expression as the destination; did you mean to provide an explicit length?

Note: I m using -Wall -Werror flags which convert warning to error.

There is something similar here
In a comment, someone answered this

"For fixed length buffers, I usually use strncpy(dest, src, sizeof(dest)); dest[sizeof(dest)-1] = '\0'; That guarantees NULL termination and is just less hassle than snprintf not to mention that a lot of people use snprintf(dest, sizeof(dest), src); instead and are very surprised when their programs crash arbitrarily."

But this is wrong:
gcc 4.8 say

"error: argument to ‘sizeof’ in ‘strncpy’ call is the same expression as the destination; did you mean to provide an explicit length? [-Werror=sizeof-pointer-memaccess]"

in gcc 4.8 documentation, they are talking about this issue:
they say:

The behavior of -Wall has changed and now includes the new warning flag -Wsizeof-pointer-memaccess. This may result in new warnings in code that compiled cleanly with previous versions of GCC.

For example,
include string.h

struct A { };

int main(void) 
{
    A obj;
    A* p1 = &obj;
    A p2[10];

    memset(p1, 0, sizeof(p1)); // error
    memset(p1, 0, sizeof(*p1)); // ok, dereferenced
    memset(p2, 0, sizeof(p2)); // ok, array
    return 0;
}
Gives the following diagnostic:
warning: argument to ‘sizeof’ in ‘void memset(void*, int, size_t)’ call is the same expression as the destination; did you mean to dereference it? [-Wsizeof-pointer-memaccess]
  memset(p1, 0, sizeof(p1)); // error
                       ^
Although these warnings will not result in compilation failure, often -Wall is used in conjunction with -Werror and as a result, new warnings are turned into new errors.
To fix, either re-write to use memcpy or dereference the last argument in the offending memset call.*

Well, in their example, it's obvious the code was wrong, but in my case, with snprintf/strncpy, I dont see why, and I think it's a false positif error of gcc. Right ?

thanks for your help
解决方案
An array decays into a pointer to the first element when you passed to a function. So what you have in 

static inline void toto(char str[3]) {..} 

is not an array but a pointer.

Hence, gcc rightly warns.

Whether you specify the size in the function parameter or not doesn't matter as:
static inline void toto(char str[3])
and
static inline void toto(char str[])
and
static inline void toto(char *str)
are all equivalent.

Read here on this: what is array decaying?

                        
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