GATE Overflow | Programming | Test 1 | Question: 18

Question

Consider the following two C codes:

Code 1	Code 2
char *a = "hello world"; printf(a);	char a[] = "hello world"; printf(a);

• A. Both are valid C codes
• B. Code 1 is valid but not code 2
• C. Code 2 is valid but not code 1
• D. Both codes are not valid in C

Answer 1

char *a = "hello world";
printf(a);

char a[] = "hello world";
printf(a);

Both are valid. In first case, programmer is not supposed to modify the contents being pointed to by a as "hello world" is a string literal. In second case, "hello world" is copied to an array a, in the function call space and hence can be modified by the programmer. For printf case, both works the same.

PS: The first argument to printf must be a character pointer and all its characters are printed with format specifiers taken care of separately. So, we do not need %s if we want to just print a string. Avoiding it is actually better for performance reason.

Comments

but printf will not work without format specifier(%s).so ,ans should be D

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Who told it won't work? The first argument to printf must be a character pointer and all its characters are printed with format specifiers taken care of separately. So, we do not need %s if we want to just print a string. Avoiding it is actually better for performance reason.

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self dobut for 2nd code pf it will print hello or hello wolrd

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Hello World. Printf for strings always goes till "\0" while scanf for string stops at white space.

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Extra Note:->>Here sizeof(a) for code1 is 8bytes or 4 bytes(sizeof pointer) where as  for code2 it is 12 bytes(including \0).

+ http://stackoverflow.com/questions/1704407/what-is-the-difference-between-char-s-and-char-s-in-c

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Sir , how

#include<stdio.h>

int main()

{

char *a="hello world";

printf(a);

}

how will it work?

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"programmer is not supposed to modify the contents being pointed to by a as "hello world" is a string literal." Can you please explain it more? Does that mean if a pointer points to a literal, then changing the literal using indirection operator on pointer can cause problem?

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IN STATEMENT 2

A is just mnemonic , it is not variable .so how can we print(A) ??

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while printing, how printf will know that it have to print as %c or %s. I mean how printf will know what to print - string or a character.

because in both case a is pointing to a single character(1st character)

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First argument to printf is const char * -- means it is always treated as a string.

Answer 2

option (a) is correct

because pointer and array both are similar in a language .

the diffrence is that pointer is a variable and array is a newmonic .

both declration is correct

Answer 3

In C, arrays and pointers are identical in the way they access the memory.

Hence, Option A

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The major difference, however, is that the pointer identifier can take different values if we assign them to it. But the name of the array points to the base address of the array, and it can't be changed.

In plain English, a pointer A is a variable, while the array name A would be a const int*.