读取二进制文件，将2字节转换为无符号短C

Question

I have a problem with reading 2 bytes at onces and convert it to an unsigned short, big endian.

This is my current code, I want to print the unsigned short big endian as well, and it should be the number 25.

So this code is for reading a binary file, I saved all the files to a buffer, and I need buffer[5] and buffer[6] to the unsigned short, big endian

void read_binair(const char* filename)
    {
        FILE *file;
        char *buffer;
        unsigned long fileLen;
        int i;
        char character;
        Level* level = level_alloc_empty();

        file = fopen(filename, "rb");
        if (!file)
        {
            fprintf(stderr, "Unable to open file %s", filename);
            return;
        }



        fseek(file, 0, SEEK_END);
        fileLen = ftell(file);
        fseek(file, 0, SEEK_SET);


        buffer = (char *)malloc(fileLen + 1);
        if (!buffer)
        {
            fprintf(stderr, "Memory error!");
            fclose(file);
            return;
        }



        fread(buffer, fileLen, 1, file);
        for (i = 0; i < 4; i++) 
        {   
            printf("%c", (char) buffer[i]);
        }
        i = buffer[4];
        printf("%d", i);
        //read buffer[5] and buffer[6] together as a unsigned short, big endian
        fclose(file);
    }

Answer 1

以下代码将在big endian中从buffer创建一个unsigned short ：

unsigned short us = (buffer[5] << 8) | buffer[6];

Answer 2

Assuming -

1. you are using a compiler which supports C99, and
2. the data in the file is in network byte order

You need -

uint16_t half_word = ((uint8_t)buffer[5] << 8) | ((uint8_t)buffer[6])

PS -

• Use uint8_t *buffer instead of char *buffer to obviate the need for typecasting buffer octets.
  • See http://pubs.opengroup.org/onlinepubs/009695399/basedefs/stdint.h.html
• If you are using C89/90 -
  • typedef unsigned char uint8_t;
  • typedef unsigned short int uint16_t;