请解释此代码执行的操作。 [英] Please explain what this code performs.

问题描述

有人请解释这段代码的作用...我正在研究音频隐写

for（i = 24，j = 0; i> = 0; i- = 8，j ++）

{

tempInt = inputFileSize;

tempInt>> = i;

tempInt& = 0x000000FF;

tempByte [j] =（byte）tempInt;

}



我的尝试：



为什么循环从24开始？

Someone please explain what does this code do ... i m working on audio steganography
for(i=24, j=0; i>=0; i-=8, j++)
{
tempInt= inputFileSize;
tempInt>>= i;
tempInt&= 0x000000FF;
tempByte[j]= (byte) tempInt;
}

What I have tried:

why the loop starts from 24??.

推荐答案

代码分裂了32位数字 inputFileSize 以相反的顺序存储到 tempByte [] 的4个字节。它类似于

The code is splitting the 32-bit number inputFileSize into 4 bytes stored to tempByte[] in reverse order. It is similar to

tempByte[0] = (inputFileSize >> 24) & 0xff;
tempByte[1] = (inputFileSize >> 16) & 0xff;
tempByte[2] = (inputFileSize >> 8) & 0xff;
tempByte[3] = inputFileSize & 0xff;

循环从24开始，因为这是最高字节必须移位多少位才能使其进入8位值的位置。

The loop starts at 24 because that is by how many bits the highest byte must be shifted to get it into the position of an 8-bit value.

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调试器在这里向您展示您的代码正在做什么，您的任务是与它应该做什么进行比较。

调试器中没有魔法，它没有找到错误，它只是帮助你。当代码没有达到预期的效果时，你就会接近一个错误。

When you don't understand what your code is doing or why it does what it does, the answer is debugger.
Use the debugger to see what your code is doing. Just set a breakpoint and see your code performing, the debugger allow you to execute lines 1 by 1 and to inspect variables as it execute, it is an incredible learning tool.

Debugger - Wikipedia, the free encyclopedia[^]
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The debugger is here to show you what your code is doing and your task is to compare with what it should do.
There is no magic in the debugger, it don't find bugs, it just help you to. When the code don't do what is expected, you are close to a bug.

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