我应该怎么做才能通过命令行C程序的整个返回值？ [英] What should I do to get the whole return value of c-program from command line?

问题描述

我有一个简单的C程序./ my_program

I have a simple C-program "./my_program"

#include <stdio.h>

int main (int argc , char **argv) {


   unsigned int return_result = 0x474;
   printf("return_result = %d = 0x%x \n",return_result,return_result);
   return return_result;

}

其结果是，该程序打印：

As a result, this program prints:

return_result = 1140 = 0x474 

我想获得C程序的bash脚本中的返回值。
根据这个链接
<一href=\"http://stackoverflow.com/questions/8230265/anyway-to-get-return-value-of-c-program-from-command-line\">Anyway从命令行得到C程序的返回值？
我应该从$得到这个变量？

I want to get a return value of c-program within bash script. According to this link Anyway to get return value of c program from command line? I should get this variable from $?

但是，当我推出这样的命令造成的：

But when I launch such commands consequence:

./my_program
echo $?

我得到

116

这是显而易见的，即116 = 0x74（最低字节）。
但是，我想对整个事件无符号整型值。
哪里不对？我应该怎么做才能通过命令行C程序的整个返回值？这是不是只是无符号整型类型。我应该在的情况下做，如果我创造了一些复杂的类型，例如， - 结构

It is obviously, that 116 = 0x74 (the lowest byte). But I'd like to get the whole unsigned int value. What is wrong? What should I do to get the whole return value of c-program from command line? This is not about only "unsigned int" type. What should I do in case if I created some complicated type, for example, - structure.

我试过这样：

return ((unsigned int) return_result_section_size);

这是行不通的。

推荐答案

退出codeS在Unix上被限制在单字节。如果你想输出更多，你可以把它写到标准输出或不是文件。

Exit codes on Unix are restricted to a single byte. If you want to output more, you could write it to stdout or to a file instead.

那么为什么C标准法令，主收益 INT ，而不是字符？我不知道...

Then why does the C standard decree that main returns int and not char? I have no idea...

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