Creating a pointer to an array of structs

This is a discussion on Creating a pointer to an array of structs within the C Programming forums, part of the General Programming Boards category; Hello, I was wondering if someone could please point out the incorrect use of syntax below, I want to create ...

  1. #1
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    Creating a pointer to an array of structs

    Hello, I was wondering if someone could please point out the incorrect use of syntax below, I want to create a pointer to an array of structs and then malloc that amount of memory....I currently have:


    Particle *pParticle[MAX_NUMBER_PARTICLES];

    pParticle = (Particle*)(malloc(MAX_NUMBER_PARTICLES * sizeof(Particle)));


    Should I use a type of loop to malloc each array element in turn instead?

    Thanks again.

  2. #2
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    Two options

    First one

    Particle *pParticle[MAX_NUMBER_PARTICLES];

    for(i=0;i<MAX_NUMBER_PARTICLES;i++)
    pParticle[i] = (Particle*)(malloc(sizeof(Particle)));

    second one

    Particle *pParticle;

    pParticle = (Particle*)(malloc(MAX_NUMBER_PARTICLES * sizeof(Particle)));

  3. #3
    C++ Developer XSquared's Avatar
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    But an array is a pointer... so it would just be:

    Code:
    Particle **pParticle;
    ( *pParticle ) = malloc( sizeof( Particle * MAX_NUM_PARTICLES ) );
    Naturally I didn't feel inspired enough to read all the links for you, since I already slaved away for long hours under a blistering sun pressing the search button after typing four whole words! - Quzah

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  4. #4
    Just Lurking Dave_Sinkula's Avatar
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    >But an array is a pointer

    The penalty for repeating such heresy is to review this.
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    King of the Internet Fahrenheit's Avatar
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    Originally posted by Dave_Sinkula
    >But an array is a pointer

    The penalty for repeating such heresy is to review this.
    Code:
    char abc[5] ;
    abc POINTS to statically allocated memory, correct?

  6. #6
    Code Goddess Prelude's Avatar
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    >( *pParticle ) = malloc( sizeof( Particle * MAX_NUM_PARTICLES ) );
    This is incorrect. The multiplication should not be an expression for the sizeof operator. This is what you want:
    Code:
    malloc( sizeof( Particle ) * MAX_NUM_PARTICLES );
    Let's move to the other side of the assignment operator for a moment:
    Code:
    ( *pParticle )
    What does pParticle point to? If you said "Nothing predictable, and certainly nothing this program owns" you get a cookie.

    >But an array is a pointer...
    ...when used as an expression.
    My best code is written with the delete key.

  7. #7
    C++ Developer XSquared's Avatar
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    Sorry 'bout that. I missed the closing parentheses on the sizeof( ) statement.

    So how would you do it? My way was just a guess.

    Would you have to do pParticle = malloc( sizeof( Particle * ) ) first?
    Naturally I didn't feel inspired enough to read all the links for you, since I already slaved away for long hours under a blistering sun pressing the search button after typing four whole words! - Quzah

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  8. #8
    Code Goddess Prelude's Avatar
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    >Would you have to do pParticle = malloc( sizeof( Particle * ) ) first?
    It depends on what you want. If you want a dynamic array of pointers to Particle (as seems likely) then you would do this:
    Code:
    pParticle = malloc ( N * sizeof *pParticle );
    Where N is the number of items you want in the array. You can use sizeof ( Particle * ), but that tends to confuse people, so taking the size of *p where p is the pointer being assigned to is better. It also improves maintainablility.
    My best code is written with the delete key.

  9. #9
    and the hat of int overfl Salem's Avatar
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    > abc POINTS to statically allocated memory, correct?
    No it doesn't point to the memory, it is the memory allocated to the array, and abc is the name by which those 5 chars are known.

    In some contexts, the name "abc" gets converted into "a pointer to the first element", but that doesn't affect the array itself - for example
    strcpy( abc, "yes" );
    But then you could also write
    strcpy( &abc[0], "yes" );

    int foo = 2;
    foo is the memory location containing 2, not a pointer to a memory location containing 2.
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