15

假设我有一个二进制文件;它包含正二进制数,但以小端写成 32 位整数

我如何阅读这个文件?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;
}
4

3 回答 3

20

这是一种适用于大端或小端架构的方法:

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;
}
于 2012-10-21T19:36:09.240 回答
12

如果您使用的是 linux,您应该看这里;-)

它是关于有用的功能,例如le32toh

于 2012-10-21T19:46:57.887 回答
5

来自CodeGuru

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

因此,您可以直接读取unsigned int然后调用它。

while ( fread(&num, 1, 4,fp) != 0) {
    endian_swap(num); 
    // conversion done; then use num
}
于 2012-10-21T19:33:47.697 回答