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address of pointer

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# 15  
Old 08-16-2006
Quote:
Originally Posted by Poison Ivy
this is still not working Smilie i get an empty line
at the top *firstFree = heap is suppose to make heap point at the first value right? then later on i change the address of firstFree so it points at first free memory space.. so i'm wanting that address to use in the for loop

i'm not sure what im doing anymore Smilie
# 16  
Old 08-16-2006
Try to dump your program and post here what values have firstFree and heap (use gdb or so) in dumpHeap()
# 17  
Old 08-16-2006
Quote:
Originally Posted by Hitori
Try to dump your program and post here what values have firstFree and heap (use gdb or so) in dumpHeap()
im kinda new to this so umm what do u mean Smilie
# 18  
Old 08-16-2006
Compile your programm with gcc -g (with debug info) and then run it with gdb (gdb programm). Look at man gdb or http://www.gnu.org/software/gdb/documentation/ for more info
# 19  
Old 08-16-2006
char heap[ 134 ];
char *firstFree = heap;

This is your first two lines of code
heap - is a character array
firstFree - is a character pointer which can hold an address of a variable

heap - holds the address of the first location of the array heap[134]

so the second line must be

firstFree=heap;

Regards
kingskar
# 20  
Old 08-16-2006
Quote:
Originally Posted by kingskar
char heap[ 134 ];
char *firstFree = heap;

This is your first two lines of code
heap - is a character array
firstFree - is a character pointer which can hold an address of a variable

heap - holds the address of the first location of the array heap[134]

so the second line must be

firstFree=heap;

Regards
kingskar
No,
char *firstFree = heap;
equals to
char *firstFree;
firstFree = heap;

It's just a combination of declaration of pointer to the charachter and it's definition.
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