a problem about malloc()


 
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# 1  
Old 08-05-2002
a problem about malloc()

1 . Thanks everyone who read the post.

2 . the programe is that :
#include <stdio.h>
#include <string.h>

void do_it(char *p)
{
p = (char *) malloc(100);

(void )strcpy(p,"1234");
}

int main(void)
{
char *p;

do_it(p);

(void )printf("p = %s \n",p);

return 0;
}

cc -g a.c -o a.out
./a.out
it will show that "memory fault(coredump)"

3 . Why ? I think that malloc() get memory from heap , and the heap can be seen at everywhere in the programe .In other words , it is "global" .

4 Then I write another programe. it works correctly .

#include <stdio.h>
#include <string.h>

char *do_it(void)
{
return ((char *) malloc(100));
}

int main(void)
{
char *p;

p = do_it();

strcpy(p, "asdfa");

(void) printf("p = %s \n",p);
}

5 . Then , the point 4 declare that point 3 is correct , but it is not we want .
6 . Then I think the OS has changed the value p when do_it() function return
# 2  
Old 08-05-2002
1 .I found the answer .
2 .First , I write another programe .
#include <stdio.h>
#include <string.h>

int main(void)
{
char *p;

p = (char *) malloc(100);

strcpy(p, "asdfasdf");

return 0;
}

It works correctly .
3 . Why ? The point is that :
p is a local varible , when we call do_it(p) , the OS pass a copy of p to child function .Though we malloc memory to p in child function ,the p variable in main function is not changed .
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