memcpy with struct and pointer

This is a discussion on memcpy with struct and pointer within the C Programming forums, part of the General Programming Boards category; Hi everyone. I am trying the following. Code: //********************* typedef struct{ unsigned char Dlength; unsigned char *Value; } AttributeA; //In ...

  1. #1
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    memcpy with struct and pointer

    Hi everyone.

    I am trying the following.
    Code:
    //*********************
    typedef struct{
     unsigned char Dlength;
     unsigned char *Value;
     } AttributeA;
    
     //In the main
     unsigned char userData[] = {0x01,0x02,0x03,0x04,0x5};
     
     AttributeA buffSrc;
     AttributeA buffDst;
     buffSrc.Dlength = 5;
     buffDst.Dlength = 5;
     buffSrc.Value = userData;
    
     memcpy(buffDst.Value, buffSrc.Value, buffDst.Dlength);
     
    //**********************
    printf of buffSrc.Value gives me the output = 0x01,0x02,0x03,0x04,0x5;
    But printf of buffDst.Value does not give the expected output (as buffSrc.Value).

    I wonder what am I doing wrong?
    I have even tried
    memcpy(&buffDst.Value[0], &buffSrc.Value[0], buffDst.Dlength);

    I shall appreciate.

  2. #2
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    You have a pointer inside struct.
    You need to allocate enough memory before copying anything to it.

  3. #3
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    try at the end:

    buffDst = buffSrc;

    and it works.

  4. #4
    Master Apprentice phantomotap's Avatar
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    O_o

    Define "works"?

    Soma

  5. #5
    Reverse Engineer maxorator's Avatar
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    buffDst.Value is an uninitialized pointer, so I'm wondering why it doesn't crash for you. It has to point to some allocated memory of size at least buffDst.Dlength.
    "The Internet treats censorship as damage and routes around it." - John Gilmore

  6. #6
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    Bayint & maxorator.
    I got it.
    Thank you.

  7. #7
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    Quote Originally Posted by phantomotap View Post
    O_o

    Define "works"?

    Soma
    Code:
     unsigned char userData[] = {0x01,0x02,0x03,0x04,0x5};
    
     AttributeA buffSrc;
     AttributeA buffDst;
    
     buffSrc.Dlength = 5;
     buffSrc.Value = userData;
     fwrite(buffSrc.Value,buffSrc.Dlength,1,stdout),fflush(stdout);
    
     buffDst=buffSrc;
     fwrite(buffDst.Value,buffDst.Dlength,1,stdout),fflush(stdout);
    
     assert(5==buffSrc.Dlength && 5==buffDst.Dlength && !memcmp(buffSrc.Value,buffDst.Value,5));
    You see, no new memory needed.

  8. #8
    Reverse Engineer maxorator's Avatar
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    OP wanted to make a copy of the data. Your code does NOT make a copy of the data - it uses the same data for both structures, not 2 separate copies of it.
    "The Internet treats censorship as damage and routes around it." - John Gilmore

  9. #9
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    Code:
    printf of buffSrc.Value gives me the output = 0x01,0x02,0x03,0x04,0x5;
    But printf of buffDst.Value does not give the expected output (as buffSrc.Value).
    The question was only an expected output for buffDst like buffSrc, therefore to make an extra copy is overdesigned.

  10. #10
    Master Apprentice phantomotap's Avatar
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    You see, no new memory needed.
    O_o

    Your code is... broken? That doesn't seem to be the right word. Stupid? Crazy? Ignorant? Foolish? Dangerous?

    I'll leave that to the other philosophers, but safe to say, your "it works" code certainly does not do anything useful.

    Oh, and the source and destination "point" to the same location so the `memcmp' is pointless (probably a "NOP").



    The question was only an expected output for buffDst like buffSrc, therefore to make an extra copy is overdesigned.
    o_O

    I think I'll go with "foolish" on this one.

    Soma

  11. #11
    and the hat of wrongness Salem's Avatar
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    > The question was only an expected output for buffDst like buffSrc, therefore to make an extra copy is overdesigned.
    Try applying that logic to your file system, when you try to make a backup copy of a file.

    "It's only a copy, I'll just point at it, no need to waste disk space".

    "WTF - my backup's been trashed!"

    Do you now understand why shallow copies are a PITA?
    If you dance barefoot on the broken glass of undefined behaviour, you've got to expect the occasional cut.
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