Thread: goto

Code:

#include <iostream.h>
void main(){
  gaurav:
  int i = 0;
  cout<<"here "<<++i;
  goto gaurav;
}

/*
can someone please tell me what happens to the previous i , each time goto is executed. I know goto s are not good to use. Just asking for information. What is the memory behavior in this case ??
*/

The i is reset to 0 on each loop. Then because you are precincreminting the i it will be equal to 1 when you display it. If you were to post increment it it would display as 0.

A better, equavilant program

Code:

#include <iostream>
using namespace std;
int main(){
	while (1)
	{
	int i = 0;
	cout<<"here "<<++i;
	}
}

and sorry for my spelling today

thank for your post , but i know what happens when the program is executed. I wanted to know how memory allocation etc occur. Will it encounter a memory full ?? , Is the memory of previous i locked ??

How does the stack behave ??

there is only one instance of i. All auto variables are allocated on the stack immediately on program entry, not when encountered in the program.

u mean to say when this function is entered atleast four int memory spaces will be allocated on the stack??

Code:

void func(int a){
if(a){
int p,q;
// do anything
}else{
int r,s;
//do anything
}
}

Originally Posted by agarwaga

u mean to say when this function is entered atleast four int memory spaces will be allocated on the stack??

Code:

void func(int a){
if(a){
int p,q;
// do anything
}else{
int r,s;
//do anything
}
}

Yes -- the compiler allocates all memory for the function in one big block, and it is not deallocated until the function returns. So, in your example, if sizeof(int) == 4, then the compiler will allocate 4*4= 16 bytes for the 4 variables. It may allocate more variables than you requested, the additonal ones are reserved for the compiler's use for temporary storage.

Code:

void func(int a){
{
  int a,b;
  for( ... )
  {
     int c;
   }
   if( ... )
   {
      int c;
   }
}

above will allocate all thee variables (a, b, and c) at beginnnig of the function. Normally there is only one integer allocated for variable c because of scoping.

Originally Posted by Wraithan

A variable is initializied when its scope is activated I thought.

You are correct -- it is initialized when the variable comes into scope but no additional memory is allocated for it. The same applies to C++ classes -- initialized (constructor called) only when they come into scope but actual allocation on the stack occurs upon function entry.

Originally Posted by agarwaga

u mean to say when this function is entered atleast four int memory spaces will be allocated on the stack??

Maybe? It depends on what your compiler does. Compilers can do whatever they want. Your function here could use no extra memory on the stack, keeping everything inside the processor, depending on the code involved. I would be disappointed if the compiler allocated 4*4 bytes for p, q, r, s, because only two are ever needed at any one time. If your code has some complicated arithmetic expression in it, the compiler might need to use extra stack space to store temporary results. If you have any function calls, stack space will be used to pass arguments to the function.

By the same token, it could optimise them out of existence altogether - there is no general way of looking at a bit of code and deciding how much stack space a function needs.

Additionally, if you have a mix of local variable types, then there may be padding between them which cannot be detected from simply looking at the code. char arrays for example are often rounded up to multiples of 4 or 8 to preserve alignment of other data on the stack.