检查溢出的最快方法？

发布于 2024-07-15 09:04:01 字数 769 浏览 4 评论 0原文

这是我的尝试。关于更好的解决方案有什么建议吗？：

// for loop to convert 32 to 16 bits
uint32_t i;
int32_t * samps32 = (int32_t *)&(inIQbuffer[0]);
int16_t * samps16 = (int16_t *)&(outIQbuffer[0]);
for( i = 0; i < ( num_samples * 2/* because each sample is two int32 s*/ ); i++ ) {
    overflowCount += ( abs(samps32[i]) & 0xFFFF8000 ) ? 1 : 0; 
    samps16[i] = (int16_t)samps32[i];
}

// Only report error every 4096 accumulated overflows
if( ( overflowCount & 0x1FFF ) > 4096 ) {
    printf( "ERROR: Overflow has occured while scaling from 32 "
            "bit to 16 bit samples %d times", 
            overflowCount );
}

这是实际检查溢出的部分：

overflowCount += ( abs(samps32[i]) & 0xFFFF8000 ) ? 1 : 0;

原文

Here's my attempt. Any tips on a better solution?:

// for loop to convert 32 to 16 bits
uint32_t i;
int32_t * samps32 = (int32_t *)&(inIQbuffer[0]);
int16_t * samps16 = (int16_t *)&(outIQbuffer[0]);
for( i = 0; i < ( num_samples * 2/* because each sample is two int32 s*/ ); i++ ) {
    overflowCount += ( abs(samps32[i]) & 0xFFFF8000 ) ? 1 : 0; 
    samps16[i] = (int16_t)samps32[i];
}

// Only report error every 4096 accumulated overflows
if( ( overflowCount & 0x1FFF ) > 4096 ) {
    printf( "ERROR: Overflow has occured while scaling from 32 "
            "bit to 16 bit samples %d times", 
            overflowCount );
}

Here's the part that actually checks for overflow:

overflowCount += ( abs(samps32[i]) & 0xFFFF8000 ) ? 1 : 0;

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手长情犹 2024-07-22 09:04:01

我个人更喜欢使用 SafeInt 类来进行溢出检查。它减少了对繁琐的错误检查的需要，并将其变成易于处理但难以忽略的异常。

http://blogs.msdn。 com/david_leblanc/archive/2008/09/30/safeint-3-on-codeplex.aspx

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偷得浮生 2024-07-22 09:04:01

您已经做的事情最接近单次施法的最快速度。但是你可以省略一些代码

overflowCount += (abs(samps32[i]) & 0xFFFF8000) ? 1：0；

可以改为：

if(samps32[i]&0xFFFF8000)overflowCount++;

甚至更简单

if(samps32[i]>>15)overflowCount++;

这两个都同样快，并且都比你的更快。

如果您实际上对溢出计数感兴趣，您可以考虑使用 SIMD 操作处理整数数组。

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老旧海报 2024-07-22 09:04:01

您似乎正在检查 16 位加法是否溢出。分支来避免汇编代码中的分支

overflowCount += (samps32[i] & 0x8000) >> 15;

您可以通过仅使This 生成三个 ALU 操作但在代码中不产生。它可能比分支版本快，也可能不快。

It seems that you are checking for the overflow of a 16-bit addition. You can avoid branch in the assembler code by just having

overflowCount += (samps32[i] & 0x8000) >> 15;

This generates three ALU operations but no branch in the code. It may or may not be faster than a branching version.

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不寐倦长更 2024-07-22 09:04:01

位操作也是我的选择。目前我能想到的唯一更快的方法是使用内联汇编，在其中加载源操作数，在芯片上进行复制，截断和按位比较（这是伪伪代码）。

您的代码有一个问题：它违反了别名规则。您可以使用类似这样的内容：

union conv_t {
    int32_t i32;
    int16_t i16;
};

然后您可以确保 IQBuffer 属于该类型。最后，您可以运行：

for( i = 0; i < (num_samples * 2); i++ ) {
    <test goes here>
    samps [i].i16 = static_cast<int16_t>(samps [i].i32);
}

edit: 根据您的编辑 (https://stackoverflow.com/revisions/ 677427/list）你几乎使我的整个帖子无效。感谢您在问题中没有提及您的编辑。

Bit ops would be my choice, too. the only faster way I can imagine at the moment is to use inline assembly where you load the source operand, make a copy onboard the chip, truncate, and bitwise compare (that was pseudo pseudo code).

Your code has an issue: It violates aliasing rules. You could use something like this instead:

union conv_t {
    int32_t i32;
    int16_t i16;
};

Then you could ensure that IQBuffer is of that type. Finally, you could run:

for( i = 0; i < (num_samples * 2); i++ ) {
    <test goes here>
    samps [i].i16 = static_cast<int16_t>(samps [i].i32);
}

edit: As per your edit (https://stackoverflow.com/revisions/677427/list) you drove nearly my whole post invalid. Thanks for not mentioning your edit in your question.

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