How local array and array index variable present in memory?

Question

I got the first problem in an interview question.But I want a proper explanation for the question.I try this in home and some other confusions are also rises.

#include <stdio.h>
int main()
{
  int arr[4]={10,20,30,40};
  int i;
  for(i=0;i<=4;i++)
  printf("%d,",arr[i]);
  printf("\n");
  return 0;
}

OUTPUT
10,20,30,40,4,

The last out put was 4.but it is out of array index.Again I think that in memory variable i present after array elements.So I get this answer.

But again i confuse with this

 #include <stdio.h>

int main()
 {
   char arr[4]={10,20,30,40};
   int i;
    for(i=0;i<=4;i++)
       printf("%d,",arr[i]);
    printf("\n");
   return 0;
 }
OUTPUT
10,20,30,40,0,

Again more confuse with below

#include <stdio.h>
int main()
 {
   int arr[4]={10,20,30,40};
   char i;
   for(i=0;i<=4;i++)
      printf("%d,",arr[i]);
   printf("\n");
   return 0;
}

OUTPUT
10,20,30,40,74743796,

Can any body Explain Why this type of variation in output?

I use intel cpu,Ubuntu os,Gcc complier..

If compiler specific or architecture specific then also mention in the answer please.

It is tricky to interpret the results of undefined behavior. — jxh
– jxh, Commented Aug 7, 2012 at 7:14
This is technically undefined behavior but its actually easy to see what is happening here. I will post more later if someone else has not — Swiss
– Swiss, Commented Aug 7, 2012 at 7:19

user703016 · Accepted Answer · 2012-08-07 07:15:30Z

1

This is called Undefined Behavior. Since you are accessing an array outside of its bounds, anything can happen and results do not have to (nor they will) make sense.

answered Aug 7, 2012 at 7:15

user703016

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Comments

mathematician1975 · Accepted Answer · 2012-08-07 07:15:53Z

1

You should not access out of bounds array memory. The value of out of bounds array access could be anything - there is no expected value for it and it has undefined behaviour. It seems like you are expecting the char and int variables that you create locally after the array to be located in memory after the array variable. Even if this were the case you have not initialised these variables and so their value could be anything. Simply don't access out of bounds memory and certainly dont try to predict the results of accessing it.

answered Aug 7, 2012 at 7:15

mathematician1975

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1 Comment

rajesh6115 Over a year ago

then what is the proper ans for this type of questions?just undefined behavior...or some thing regarding memory allocation of local variable.

Blagovest Buyukliev · Accepted Answer · 2012-08-07 07:29:52Z

What you are dealing with is undefined behavior.

In the first case it just happens that the i variable is placed right after the arr array on the stack, so getting out of the bounds of the array you are touching the i variable. The types of i and arr are the same (int), so printing out i as if it were a part of the array is OK.

In the second case you are trying to access i as if it was a char variable, so you only get a part of it which is zero.

Have in mind that the compiler is free to rearrange the variables on the stack, so any assumption that the i variable should follow right after (even though it is declared in that order) is incorrect.

To the compiler, arr and i are just two separate local variables. The size of arr is not kept anywhere in memory: arr is just a piece of memory with a starting address. You can examine its size at compile time via the static sizeof operator.

Community · Accepted Answer · 2020-06-20 09:12:55Z

0

To be clear, what you are seeing is classified as "Undefined Behavior", but there is enough information in your answer to know what is going on.

First, some detail on what we can see from your question about the architecture of your machine.

About the CPU

CPU is little endian

About the OS

Stack grows downwards

About the Compiler

int is 32 bit in size
char is 8 bit in size
Local variables are ordered on the stack the same as they are in the code
Puts char into larger block than it needs to speed up run time
Uses memory padding to speed up code

All of these things combine to end up with the result you are seeing. A visual demonstration is best for showing what happens from here.

The First Case

At the end of the loop, the stack looks like this:

| 10 | 0 | 0 | 0 | a[0]
| 20 | 0 | 0 | 0 | a[1]
| 30 | 0 | 0 | 0 | a[2]
| 40 | 0 | 0 | 0 | a[3]
|  4 | 0 | 0 | 0 | i (a[4])

The value for i is in the same spot that a[4] should be. So a[4] has the same value as i.

The Third Case

What i looks like depends on a lot of variables. But it seems like this is the layout that makes the most sense.

| 10 | 0 | 0 | 0 | a[0]
| 20 | 0 | 0 | 0 | a[1]
| 30 | 0 | 0 | 0 | a[2]
| 40 | 0 | 0 | 0 | a[3]
|  X | X | X | 4 | i (a[4])

This is reflected in the value you see printed. 74743796 in little endian binary is

11110100 01111111 01110100 00000100

The last byte there is 4. When you reference i in the code, the context is as a char and only the last byte is used. When you reference a[4], the context is as an int and all four bytes are used.

This is important because it shows that the compiler is shoving chars into the last byte and padding the rest with garbage.

The Second Case

Based on the other two, I would expect the memory to be as follows

| X | X | X | 10 | a[0]
| X | X | X | 20 | a[1]
| X | X | X | 30 | a[2]
| X | X | X | 40 | a[3]
| 4 | 0 | 0 |  0 | i (a[4])

The magic comes into play with the last memory line. When it is accessed as an int, it is processed normally. However, when it is accessed as a char, the compiler assumes the first three bytes are padding and returns the last byte, or 0

edited Jun 20, 2020 at 9:12

CommunityBot

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answered Aug 7, 2012 at 8:20

Swiss

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3 Comments

rajesh6115 Over a year ago

for second case explanation you violate array concept (that is array elements present at consecutive location only).But from your explanation I get an idea to conform this things.

Swiss Over a year ago

@Rajesh I did overlook that. The array would be | 10 | 20 | 30 | 40 | in that case. It does also complicate the fact that a[4] would look at the first byte of i. Maybe the CPU is big endian?

rajesh6115 Over a year ago

I use Intel,Ubuntu and Gcc compiler,And My machine is little endian.And all are my system outputs ...

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