umesh_singh

 Profile Answers by umesh_singh 

• Mar 26th, 2007

 

it will allocate 200 bytes of memory and returns pointer to the first location...

correct me if iam wrong


regards
umesh

sourav punoriyar

 Profile Answers by sourav punoriyar 

• Sep 5th, 2013

 

Code
int *ptr = (int *)malloc(100*(sizeof(int)));


ptr++;


free(ptr);


 

i think it will allocate 400 bytes...int taken as 4 bytes..gcc compiler..and send base address to pointer ptr...
next ptr is incremented ptr++=ptr+4..points to next location

now free(ptr)...this will free the memory but ptr will still point to the previous memory location..as free return type is void...
so we have to write ptr=0;
to make it clear..
output no may be same if os doesn't use those mem location used by us previously..and may be different ..depends on o.s.
plz correct if wrong

jbode

 Profile Answers by jbode Questions by jbode

• Nov 29th, 2013

 

First of all, you can clean up that malloc call (see attached code). You shouldn't cast the result of malloc; its unnecessary, and under compilers that use the 1989/1990 standard, it can suppress a diagnostic if you dont have a declaration for malloc in scope.

The result will likely be a runtime error, since pointer value being passed to free is not the same value that was returned from malloc.

Code
int *ptr = malloc(100 * sizeof *ptr);