C：如何摆脱转换错误？

问题描述

我有一个使用gcc版本4.6.3的项目，我不得不使用-Wall -Werror -Wconversion进行编译。下面的简单示例显示了我无法摆脱的错误：

  #include  
 
 int main（void）{
 uint32_t u = 0; 
 char c = 1; 
 
 u + = c; 
 return（int）u; 
 
 
 
 $ b 
用上面的标志编译它：
  test.c：7：8：错误：从'char'转换为'uint32_t'可能会改变结果的符号[-Werror = sign-conversion] 
  
好的，很好。只需添加一个类型转换，对吧？不。将第7行更改为 u + =（uint32_t）c 不会使错误消失。即使将它改为 u = u +（uint32_t）c 也不会让它消失。
 
 
 
是否有可能解决这个问题？
 
 
 
请注意，char来自一个字符串，所以我没有选择改变它的类型。 b $ b 
解决方案
这里编译得很好： 
 
 
  u + =（unsigned char）c; 
  
然而，这只会使警告无效  - 不会对每个人执行任何操作在运行时 c ，与Basile的建议不同。
 
I have a project which uses gcc version 4.6.3, and I'm forced to compile with "-Wall -Werror -Wconversion".  The following simple example shows an error I can't get rid of:
#include <stdint.h>

int main(void) {
  uint32_t u = 0;
  char c = 1;

  u += c;
  return (int)u;
}
Compiling it with the above flags gives:
test.c:7:8: error: conversion to ‘uint32_t’ from ‘char’ may change the sign of the result [-Werror=sign-conversion]
Ok, fine.  Just add a typecast, right?  Nope.  Changing line 7 to u += (uint32_t)c does not make the error go away.  Even changing it to u = u + (uint32_t)c does not make it go away.
Is it possible to fix this?
Please note that the "char" is coming from a string, so I don't have the option to change its type.
 解决方案 
This compiles fine here:
u += (unsigned char)c;
This will only silence the warning, however — without doing anything to each c at run-time, unlike Basile's proposal.
                        
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