从字节的char []中获取int [英] Get int from char[] of bytes

问题描述

我有一个读入char数组的文件.

I have a file that I read into a char array.

char数组现在拥有一定数量的字节，如果我知道文件在小字节序存储的0x8位置存储了一个32位整数，该如何从char数组中检索它?

The char array now holds a certain number of bytes, now if I know the file stored a 32bit integer in little endian at position say 0x8, how do I retrieve it from the char array?

FILE * file = fopen("file");
char buffer[size];
fread(buffer,1,size,file);
int = buffer[0x8]; // What should I do here?
// I imagine it involves some strange pointer
// arithmetic but I can't see what I should do,
// casting to int just takes the first byte,
// casting to (int *) doesn't compile

推荐答案

您需要像这样进行投射:

you need to cast it like so:

int foo = *((int *) &buffer[0x8]);

首先将点转换为int指针，然后将其取消引用为int本身.

Which will first cast the spot to a int pointer and the dereference it to the int itself.

[但是要注意不同机器类型之间的字节顺序；有些先做高字节，有些先做低字节]

[watch out for byte-ordering across different machine types though; some do high bytes first some do low]

为了确保示例易于理解，下面的代码显示了结果:

And just to make sure the example is well understood, here's some code showing the results:

#include <stdio.h>

main() {
    char buffer[14] = { 0,1,2,3,4,5,6,7,8,9,10,11,12,13 };
    int foo = *((int *) &buffer[0x8]);
    int bar = (int) buffer[0x8];

    printf("raw: %d\n", buffer[8]);
    printf("foo: %d\n", foo);
    printf("bar: %d\n", bar);
}

运行结果以及

raw: 8
foo: 185207048
bar: 8

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