GATE IT 2004 | Question: 61

Question

Consider the following C program:

#include <stdio.h>
typedef struct {
    char *a;
    char *b;
    } t;
void f1 (t s);
void f2 (t *p);
main()
{
    static t s = {"A", "B"};
    printf ("%s %s\n", s.a, s.b);
    f1(s);
    printf ("%s %s\n", s.a, s.b);
    f2(&s);
}
void f1 (t s)
{
    s.a = "U";
    s.b = "V";
    printf ("%s %s\n", s.a, s.b);
    return;
}
void f2(t *p)
{
    p -> a  = "V";
    p -> b = "W";
    printf("%s %s\n", p -> a, p -> b);
    return;
}

What is the output generated by the program ?

• A. $A \ B$
  $U \ V$
  $V \ W$
  $V \ W$
• B. $A \ B$
  $U \ V$
  $A \ B$
  $V \ W$
• C. $A \ B$
  $U \ V$
  $U \ V$
  $V \ W$
• D. $A \ B$
  $U \ V$
  $V \ W$
  $U \ V$

Comments

In c programming whenever we pass something using "call by value" the value of the variable is passed to a new variable of same data type(whether it is an integer, pointer or structure).

Answer 1

First print $\text{A  B}$

$\textsf{f1}$ is call by value the changes applicable only for local

from $\textsf{f1},  \text{U V}$ is printed

back in main $\text{A B}$ is printed

then in $\textsf{f2}, \text{V W}$ is printed

Hence, answer is (B).

Comments

@Arjun Sir what difference will "static" make in the above program ? what will happen if it isnt there ?

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@Danish: 'static' variables retain their values b/w function calls and are NOT INITIALIZED when function is called again. Life of a static variable is THE EXECUTION TIME OF PROGRAM and NOT the function block.

But here this static DOES NOT seem to affect anything because main() is called just once. Even if it was not static it would have retained the value as main() is always last to finish.

YES, it would have made significant difference if variables in f1() or f2().

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That is why I was confused that if it is not making any difference then why is it there ... Thanks ;)

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#include <stdio.h>
typedef struct {
    char *a;
    char *b;
    } t;
void f1 (t s);
void f2 (t *p);
main()
{
    static t s = {"A", "B"};
    printf ("%s %s\n", s.a, s.b);
    printf("%u",s.a);
    f1(s);
    printf ("%s %s\n", s.a, s.b);
    f2(&s);
}
void f1 (t s)
{
    //s.a = "U";
    //s.b = "V";
    printf("\n%u\n",s.a);
    printf ("%s %s\n", s.a, s.b);
    s.a = "A";
    s.b = "V";
     printf("\n%u\n",s.a);
    printf ("%s %s\n", s.a, s.b);
    return;
}
void f2(t *p)
{
    p -> a  = "A";
    p -> b = "W";
    printf("%s %s\n", p -> a, p -> b);
    printf("\n%u",p->a);
    return;
}

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In whole program wherever we refer same string constant(i.e "A") irrespective of scope of the function we get the same base address. Why??? I think it should be different ...@Arjun

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why it should be different?

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@Arjun sir ..

Suppose in code it is given that...

If( "ABC"==" ABC" )

 Print ("hello");

Else

Printf("yes");

What is output...??

---

checkliteral.c: In function ‘main’:
checkliteral.c:4:10: warning: comparison with string literal
results in unspecified behavior [-Waddress]
 if( "ABC"==" ABC" )

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Sir..string literal returns base address...

Rt??

Like char *p = "ABC";

P will contain the base address of " ABC " ..rt??

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@gabber

Whenever a structure is passed as an argument, the whole structure is copied and given to the local structure variable of the called function, obviously the addresses of both the structures should be different. But, C language does a kind of optimization here.

The local structure variable of the called funtion will keep on pointing to the same structure unless that structure is modified, and as soon as you modify, a new structure is now created and all the contents of the previous structure are copied to the new structure now and the local structure variable will now point to a different address.

You may remove the comment operator in the function f1() in s.a = "U", and re-run your program. You will find the addresses different this time.

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Struct s stores the base addresses of strings. In function f1 struct s is passing as agrument.and struct s is copied into local struct variable s of f1() function..

This s contains same base addresses for A and B or not?

I mean why not f1( ) updating the original values stored (A and B)?

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struct variable is just like any other variable. Let the 's' in main be a box and it has two codes to lockers 'A' and 'B'. When it is passed to 'f1', a copy of the box is made and the codes are copies to it. Inside the function 'f1' the codes are changed -  the new values won't open our original locker. But once function returns, we have our original codes in the box 's'.

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Its like inside f1 function s will be copied and it will have two pointers a and b pointing to another addresses storing the A nd B initially. Right?

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@Arjun sir, we cannot modify string constant right??? Then how is f2 printing V W???

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Please explain fisrt few steps using diagram ....how actually it look like after declaration and  initialization ....

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@ ravi_ssj4 Wow!! How do you know this?

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@papesh

AFAIK I read about it somewhere. It was mentioned that if in a program same string constant appears more than once then the compiler instead of allocating memory for them separately, it allocates memory only once and wherever the string constant appears then at that place the same base address is used (this is a clever approach of saving memory). So, probably that's why whenever you print value of 'a', then same address is being printed.

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You must have read string literal not string constants . String literals are reused from literal pool . You can find this statement in book named using pointer with c author name is resse. Good luck

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wat is the difference between string constant and string literal ??

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Hi

In C, literals (of which there are only string literals) are lvalues; we can take their address. All other “literals” (numbers, characters) are constants in C; this means they are rvalues and we cannot take their address.

In languages other than C, I tend to use “constant” and “literal” interchangeably. In C, they mean different things.

// Some constants
int i = 5;        // `5` is a constant
char c = 'x';     // `x` is a constant
uint64_t j = 45;  // `45` is a constant
char * s = NULL;  // `NULL` is a constant

// Some literals (actually, just string literals)
char * str = "hello";  // "hello" is a string literal

A literal is an lvalue: an expression with an address. This is why we have “string literals” and not “string constants”. The string literal is allocated in memory; we can take its address.

A constant is an rvalue: an expression without an address. Numbers and characters are literals. Taking the address of 45 has no meaning.

Source https://jameshfisher.com/2017/01/22/c-constants-literals.html

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A string constant or literal are the same thing as far i hav studied ... A string constant cant be manipulated ... But a character array can be manipulated ...

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@Shaik Masthan

Struct t s

Here s returns base address of structure object rt ?

why in f1(t s)..here why local object of s is created and changes made there..

In case of array,,when array is passed as parameter to function ..it is always taken as pointer and no separate copy is created ..isn't this same in case of structure ??

int main(){                                            

int array[5];

f1(array)

}

int f1(int a[]){                         //here a and array will point to same location rt?
}

--------------------------------------

int main(){                                            

Struct t s;

f1(s)

}

int f1(t p){                    //here p and s points to different objects ????

}                     

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in C, struct is just like some other datatype ( actually we called it as user defined data type. )

struct t s, is just like as int k

how they passed to function ? by default using call by value

but in case of array's it is not true.

one more difference between array and structure is , assigning.

int a[10],b[10];

a=b -----------> invalid

struct t k1,k2;

k1=k2  --------> valid !

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thanks @Shaik Masthan :)

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static t s = {"A", "B"};
    printf ("%s %s\n", s.a, s.b);
    f1(s);
    printf ("%s %s\n", s.a, s.b);
    f2(&s);    
    printf ("%s %s\n", s.a, s.b); 

@Arjun @Shaik Masthan  sir

https://gateoverflow.in/3704/gate2004-it-61?show=81095#c81095

What if there is one more $printf$ after $f2(\&).$ Does it o/p $V\ \ W$ $or$ $A\ \ B$

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@Kushagra गुप्ता

The new printf you added will print V W.

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It’s because ‘→ ‘ is used with pointers and as we have passed the address of s as (&s) in function call present in main() and further ‘→’ dereferences the address location and hence updates the value with V W, while ‘.’ is used only for normal accessing of structure? @chirudeepnamini

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@Arjun Sir

I am not getting any error(but you have commented, we will get an error  (see your previous comment in this section )

)

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Try using “-Wall” parameter to gcc.

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@gatecse

What is the purpose of -Wall

 (no effect here)

Answer 2

A B
U V
A B
V W