Arm Xscale不正确的分支

问题描述

我正在开发arm xscale嵌入式系统。在这里，我看到一个不正确的分支在开关盒中被执行。



下面是我的开关盒看起来像，

 开关（val）
 {
  case   0 ：
 .... 
  break ; 
  case   1 ：
 .... 
 断裂; 
  case   2 ：
 ..... 
 断裂; 
  case   3 ：
 ..... 
 断裂; 
 默认：
 printf（ 案例值=％d \\\\ n，val）; 
  break ; 
} 

在上面的例子中，我收到零值，即val = 0.

但是cpu分支到默认情况，我打印了值，它只显示零。 案例值= 0。看起来很奇怪。

Break语句是正确的。 val是一个带符号的int（没问题）。

我试图理解为这个switch case块生成的汇编，但是我不理解生成的汇编指令中使用的逻辑。

我不认为汇编程序生成错误的程序集（可能我不知道）。是否有任何可能的硬件错误会导致MMU或TLB中出现类似问题，指令缓存奇偶校验问题或任何其他事情？如何调试此问题？

解决方案

I am working on arm xscale embedded system. Here I am seeing a incorrect branch is getting executed in switch case.

Below is my switch case looks like,

switch(val)
{
  case 0:
         ....
         break;
  case 1:
         ....
         break;
  case 2:
         .....
         break;
  case 3:
         .....
         break;
  default:
         printf("case value = %d\r\n" ,val);
         break;
}

In the above case , I am receiving value as zero i.e val = 0.
But the cpu branches to default case, where i have printed the value, which is showing zero only. "case value = 0". Looks very strange.
Break statements are proper. "val" is a signed int(no problem there).
I have tried to understand the assembly generated for this switch case block, but I don't understand the logic used in the generated assembly instructions.
I don't think wrong assembly was generated by assembler(may be I don't know).Is there any possiblity of hardware bug which will create this kind of problem like issues in MMU or TLB , instruction cache parity issues or any other thing ?How to debug this issue?

解决方案

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