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Home/ Questions/Q 8513859
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Editorial Team
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Editorial Team
Asked: June 11, 20262026-06-11T04:41:58+00:00 2026-06-11T04:41:58+00:00

I’m relatively new to C++ and I’m trying to read a proprietary file format.

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I’m relatively new to C++ and I’m trying to read a proprietary file format. I know the header format, and I’ve created a struct RTIHeader with the necessary fields.

My test code reads bytes from the file and copies them into the same memory space as the header struct, effectively reconstituting it. My problem is that every time I run the test code (just calling the constructor) I get a different value! My theory is that I do not fully understand memcpy.

struct RTIHeader decode(char* memblock){
    struct RTIHeader header;
    memcpy(&header,&memblock,sizeof(RTIHeader));
    return header;
}

RTIFile::RTIFile(const char* filename){
    // open the file in binary input mode with the pointer at the end
    std::ifstream file(filename,
                       std::ios::in |
                       std::ios::binary |
                       std::ios::ate);

    std::ifstream::pos_type size;
    char * memblock;
    RTIHeader header;

    // if the file didn't open, throw an error
    if(!file.is_open()){
        //TODO learn about C++ error handling
        return;
    }
    // use pointer position to determine file size
    size = file.tellg();
    // read file
    memblock = new char [sizeof(RTIHeader)];
    file.seekg(0,std::ios::beg);
    file.read(memblock,sizeof(RTIHeader));

    header = decode(memblock);

    std::cout << (unsigned int)header.signature[0] << "\n";

    // still need to read the rest of the file
    file.close();
}
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  1. Editorial Team
    Editorial Team
    2026-06-11T04:42:00+00:00Added an answer on June 11, 2026 at 4:42 am

    You are passing the address of memblock as the second argument of memcpy. Since memblock is a parameter to the function, you will just be copying chunks of memory off your stack.

    To fix, just pass the pointer directly:

    memcpy(&header,memblock,sizeof(RTIHeader));
    

    By the way, since you are returning the structure by value, you will incur an extra memcpy as the returned value is copied in full (unless your compiler optimizes it). To avoid this, you should consider passing the target structure as a pointer to decode, as in

    struct RTIHeader *decode(struct RTIHeader *header, char* memblock){
        memcpy(header, memblock, sizeof(RTIHeader));
        return header;
    }
    

    Then you just have to do

    decode(&header, memblock);
    

    which is more efficient.

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