GATE IT 2006 | Question: 49

Question

Which one of the choices given below would be printed when the following program is executed ?

#include <stdio.h>
struct test {
               int i;
               char *c;
}st[] = {5, "become", 4, "better", 6, "jungle", 8, "ancestor", 7, "brother"};
main ()
{ 
    struct test *p = st;
    p += 1;
    ++p -> c;
    printf("%s,", p++ -> c);
    printf("%c,", *++p -> c);
    printf("%d,", p[0].i);
    printf("%s \n", p -> c);
}

• A. $\text{jungle, n, 8, nclastor}$
• B. $\text{etter, u, 6, ungle}$
• C. $\text{cetter, k, 6, jungle}$
• D. $\text{etter, u, 8, ncestor}$

Comments

it's always easy when u know the precedence,

but remembering the precedence is difficult, wish had known any tricks to remember it easily :)

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https://youtu.be/nKrmP4mF2NA?list=PLIPZ2_p3RNHgzQutUHGzqMjA1z7XJ_Uya&t=1452

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@usher I use this Mnemonic to remember the operator Precedence.

"All Smart Rabbits Buy Large Tasty Apple Cakes"

where,

\begin{array}{ll} A &\rightarrow \text{Arithmetic operators} \\ S &\rightarrow \text{Shift operators} \\ R &\rightarrow \text{Relational operators} \\ B &\rightarrow \text{Bitwise operators} \\ L &\rightarrow \text{Logical operators} \\ T &\rightarrow \text{Ternary operator} \\ A &\rightarrow \text{Assignment operators} \\ C &\rightarrow \text{Comma operator} \end{array}

Answer 1

code :

#include <stdio.h>

struct test {
    int i;
    char *c;
}st[] = {5, "become", 4, "better", 6, "jungle", 8, "ancestor", 7, "brother"};

int main () { 
    //printf("size = %d\n",sizeof(struct test) );
    struct test *p = st;
    p += 1;
    ++p->c; // ++(p->c)
    printf("%s,", p++->c); // (p++)->c 
    printf("%c,", *++p->c); // *(++(p->c))
    printf("%d,", p[0].i);
    printf("%s \n", p->c);
}

We will assume few things:

• Size of integer $4$ Bytes.
• Size of a pointer $4$ Bytes.

Neglecting any alignment issues with the storage of this structure we will have $8$ Bytes per structure. 

And one precedence rule we need to use:

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Initial situation :

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struct test *p = st;

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p += 1;

We know that if ptr is a pointer then,  ptr + x = ptr + x*sizeof(*ptr);

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++p->c;

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printf("%s,", p++->c);         // (p++)->c

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printf("%c,", *++p->c);     // *(++(p->c))

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printf("%d,", p[0].i);

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printf("%s \n", p->c);

Correct Answer: $B$

Comments

Whatt a explaination !!!

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thanks for such a beautiful explanation

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Oh man, what an explaination! Thank you!!

Answer 2

#include <stdio.h>
struct test {
               int i;
               char *c;
}st[] = {5, "become", 4, "better", 6, "jungle", 8, "ancestor", 7, "brother"};
main ()
{ 
    struct test *p = st;
    

p += 1;

++p -> c;
printf("%s,", p++ -> c);
    

Prints value and  increments the pointer. Pointing ti the next structure

printf("%c,", *++p -> c); // *(++( p->c ))

printf("%d,", p[0].i); //Prints the integer value associated with struct pointed by p

printf("%s \n", p -> c); //Prints the character string pointed by p

PS : Thank you @monanshi for the explanation

Comments

great explanation.......

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thor It is incrementing to the next structure in the array as p is the pointer to struct test

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what a beautiful explaination this would be my the best question among all c programming

Answer 3

Answer is B )

Run The code.

http://codepad.org/WmSREd4G

For extra inofrmation =>

++p -> c will become ++(p -> c)

p++ -> c will become (p++) -> c

 *++p -> c will become => *++(p->c)

Reference-> C Puzzle Book

Comments

in case of pre increment on p , ---> operator is given more priority  by applying brackets

in case of post increment on p , ----.> operator is given less priority  by applying brackets on p++       

 

please explain why ???/

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-> is not given priority in post inc, () is there to show ++ is associated with p not with p->c

Answer 4

I am getting B.

 

1.#include <stdio.h>
2.struct test {
3.               int i;
4.              char *c;
5.}st[] = {5, "become", 4, "better", 6, "jungle", 8, "ancestor", 7, "brother"};
6.main ()
7.{ 
8.   struct test *p = st;
9.    p += 1;
10.    ++p -> c;
11.    printf("%s,", p++ -> c);
12.    printf("%c,", *++p -> c);
13.    printf("%d,", p[0].i);
14.    printf("%s \n", p -> c);
15.}

Line 8 => Initially p is pointing to st, i.e, first element of st which is {5, "become"}
Line 9 => Now p is pointing to {4, "better"}
Line 10=> ++(p->c) since -> has higher precedence. So, p->c points to 'e' of 'better'
Line 11=> prints 'etter' and p now points to {6, "jungle"}
Line 12=> *++(p->c) since -> has higher precedence. prints 'u'
Line 13=> p->i, which is 6 so prints '6'
Line 14=> prints 'ungle' since p is pointing to 'u'.

So, output is "etter, u, 6, ungle" that is B.

Comments

@monanshi your explanation has errors. p is a pointer to structure test and it can only point to structures of the same type, so p will never point to 'e', rather p->c points to 'e' after ++p->c, same thing for line 14, when p->c points to u, not p.

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++p points to one letter of word 'e' so how can p++ point directly to next element of structure?

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p is a pointer to structure. So, why ++(p ->c) increments p by 11B. Why it doesn't increment the value of character.

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Nice Explanation Monashi. But I have added few steps

Initially st => [{5, c0},{4,c1},{6,c2},{8,c3},{7,c4}]

Suppose address of variable c is where ci are:

c0 => "become"
c1 => "better"
c2 => "jungle"
c3 => "ancestor"
c4 => "brother"

Line 8 => Initially p is pointing to st, i.e, first element of st which is {5, c0} 
Line 9 => Now p is pointing to {4, c1} 
Line 10=> ++(p->c) since -> has higher precedence. [ Means: p->c= (p->c)+1 Therefore address of c1 is incremented by 1 and thus c1 => "etter"] 
Line 11=> prints 'etter' [because c1 => "etter" now ] and p now points to {6, "jungle"} 
Line 12=> *++(p->c) since -> has higher precedence. [ Means: ++(p->c) => (p->c= (p->c)+1) Therefore address of c2 is incremented by 1 and thus c2 => "ungle"  and * (++(p->c)) prints the character pointing to, i.e 'u' ] 
Line 13=> p->i, which is 6 so prints '6' 
Line 14=> prints 'ungle' [Because of modification in the string c2 at Line 12]

Post Execution st => [{5, "become"},{4,"etter"},{6,"ungle"},{8,"ancestor"},{7,"brother"}]

Answer 5

Ans is option b

Comments

@gateasp17

can u tell why there is different memory locations for s[0] s[1] ?

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First of all there is no array name s[ ].

its st[ ]  I guess.. Here I assume THAT to store an address 8Byte is required and int requires 2 byte..  which makes st[0] location 6000 then st[1] location 6010 .

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it was typo it is st right not s

Here I assume THAT to store an address 8Byte?what does it signify

whom address u r going to store?

i compile it on my machine it give successive difference of 8  in memory address

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I mentioned in picture assume pointer size is 8 byte as in struct one int and one char pointer so take 2Byte for int and 8 byte for char. And these sizes (int char, float) can vary machine to machine as I know.

What do you want know?  

If u get succesive difference of 8 byte, Can u tell Me how much size is require in ur machine ro store an integer and character pointer or any pointer separately??

Answer 6

answer is :

option B

etter,u,6,ungle

Comments

pretty simple logic so i am not mentioning the flow... anyone wants it then i may edit my answer.

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Gott it

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Please explain ur solution.