Question

说我有一个二进制文件;它包含正二进制数，但用 little endian 写成32位整数

如何阅读此文件？我现在有这个。

int main() {
    FILE * fp;
    char buffer[4];
    int num = 0;
    fp=fopen("file.txt","rb");
    while ( fread(&buffer, 1, 4,fp) != 0) {

        // I think buffer should be 32 bit integer I read,
        // how can I let num equal to 32 bit little endian integer?
    }
    // Say I just want to get the sum of all these binary little endian integers,
    // is there an another way to make read and get sum faster since it's all 
    // binary, shouldnt it be faster if i just add in binary? not sure..
    return 0;
}

Answer 1

这是一种适用于big-endian或little-endian架构的方法：

int main() {
    unsigned char bytes[4];
    int sum = 0;
    FILE *fp=fopen("file.txt","rb");
    while ( fread(bytes, 4, 1,fp) != 0) {
        sum += bytes[0] | (bytes[1]<<8) | (bytes[2]<<16) | (bytes[3]<<24);
    }
    return 0;
}

Answer 2

如果您使用的是Linux，则应该查看here; - ）

它是关于有用的功能，例如 le32toh

Answer 3

来自CodeGuru：

inline void endian_swap(unsigned int& x)
{
    x = (x>>24) | 
        ((x<<8) & 0x00FF0000) |
        ((x>>8) & 0x0000FF00) |
        (x<<24);
}

因此，您可以直接阅读unsigned int然后调用它。

while ( fread(&num, 1, 4,fp) != 0) {
    endian_swap(num); 
    // conversion done; then use num
}

如何在C ++中从文件中读取小端整数？

3 个答案: