09-03-2002
malloc function
Hello
This is a simple program i carried out in my machine
i dont know how it is working
#include<alloc.h>
#include<stdio.h>
mian()
{
int *p,j;
p= (int*)malloc(1);
for(j=1;j<=580;j++)
{
*p=j;
++p;
}
p=p-580;
for(j=1;j<=580;j++)
{
printf("%d",*p);
}
}
Linux 7.0
gcc compiler
when i mentioned 1 byte in malloc function how it stores 580 integers ?
it is printing from 1 to 580 numbers.
but when i specified as 581 the core gets dumped for same
malloc(1) function
what is reason behind that
how it stores 580 integers in one single byte
why it has not send an message (memory insufficeint)
please explain
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end(3C) Standard C Library Functions end(3C)
NAME
end, _end, etext, _etext, edata, _edata - last locations in program
SYNOPSIS
extern int _etext;
extern int _edata;
extern int _end;
DESCRIPTION
These names refer neither to routines nor to locations with interesting contents; only their addresses are meaningful.
_etext The address of _etext is the first location after the program text.
_edata The address of _edata is the first location after the initialized data region.
_end The address of _end is the first location after the uninitialized data region.
USAGE
When execution begins, the program break (the first location beyond the data) coincides with _end, but the program break may be reset by
the brk(2), malloc(3C), and the standard input/output library (see stdio(3C)), functions by the profile (-p) option of cc(1B), and so on.
Thus, the current value of the program break should be determined by sbrk ((char *)0).
References to end, etext, and edata, without a preceding underscore will be aliased to the associated symbol that begins with the under-
score.
SEE ALSO
cc(1B), brk(2), malloc(3C), stdio(3C)
SunOS 5.10 1 Sep 2003 end(3C)