manualVSautomatic memory allocation.

This is a discussion on manualVSautomatic memory allocation. within the C Programming forums, part of the General Programming Boards category; Hello, I was wandering when and how does the compiler do the memory allocation for custom types, so that would ...

  1. #1
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    manualVSautomatic memory allocation.

    Hello, I was wandering when and how does the compiler do the memory allocation for custom types, so that would be possible to avoid the use of malloc() and free().

    Thanks for help.
    BrownB

  2. #2
    SleepWalker tjohnsson's Avatar
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    Code:
    void
    my_func ( int n ) {
       char array[n]; // this array[] life time is that function scope and has n members allocated at runtime.
        ...
    }

  3. #3
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    perhaps I should make an example...

    Let's have a struct type:
    Code:
    typedef struct{
        int a;
        int b;
        ....
    }MyStruct
    If I declare a variable of that type in my function as follows, will it be allocated in memory by the compiler?
    Code:
    void MyFunc(void){
        
        MyStruct s;
    
        s.a=1;
        s.b=2;
        ....
    }
    or the only way is to do a manual allocation:
    Code:
    void MyFunc(void){
        
        MyStruct *s;
    
        s=malloc(sizeof(MyStruct));
        
        s->a=1;
        s->b=2;
        ....
        free(s);
    }
    The question is:
    when does a compiler 'know' how to allocate a declared variable and when doesn't it know?

    I'm a PalmOS programmer, and for some system structs there is no need to use a pointer and manually allocate the memory, while for my structures I must do manually the work. I'd like to know how does the compiler work.
    Thanks.
    BrownB

  4. #4
    SleepWalker tjohnsson's Avatar
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    int b; // memory of b exist in current scope
    int b[n]; // memory of b is allocated at runtime if n size couldn't be determined at compile time.

  5. #5
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    Ok, I finally understand!!

    So, every structure can be allocated at compile time if it contains members which size can be known at compile time=>if the sizes are computed using constants.

    If a size depends from a runtime variable/value, the structure must be allocated at runtime.

    Is this right?
    Thanks a lot for help!
    BrownB

  6. #6
    not-a-geek
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    Quote Originally Posted by BrownB
    The question is:
    when does a compiler 'know' how to allocate a declared variable and when doesn't it know?
    It doesn't. If you statically declare an array or struct larger than the stack permits it, you will get an segmentation fault.

    Code:
    MyFunc:
     	pushl	%ebp
     	movl	%esp, %ebp
     	subl	$24, %esp
     	movl	$1, -8(%ebp)
     	movl	$2, -4(%ebp)
     	leal	-8(%ebp), %eax
     	movl	%eax, (%esp)
     	call	Unknownfunction
     	leave
     	ret
    As you can see, immediate 1 and 2 are assigned to locations on the stack, right behind our not-existing parameterlist. (Ignore the Unknownfunction, it's just to prevent GCC from wiping a function without in- or output).

    Code:
    MyFunc2:
     	pushl	%ebp
     	movl	%esp, %ebp
     	subl	$8, %esp
     	movl	$8, (%esp)
     	call	malloc
     	movl	$1, (%eax)
     	movl	$2, 4(%eax)
     	movl	%eax, (%esp)
     	call	free
     	movl	%ebp, %esp
     	popl	%ebp
     	ret
    Here of course the memory is allocated dynamically, eax contains the address of the malloc'd memory, movl moves the immediates into *eax and *(eax +1) (if eax were an int pointer).
    If you need to know more about how the stack itself is "allocated", have a look at intels x86 reference manual.

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