What will be the output of printf \%d?

1. When we write printf(“\%d”,x); this means compiler will print the value of x. But as here, there is nothing after \%d so compiler will show in output window garbage value.

What will be output of following printf statement?

printf() prints the text as well as the value of the variable, constant etc. printf() returns an integer value that is the total number of printed characters including spaces, carriage return, line feed etc.

How do you find the output code?

The scanf function returns the number of input is given. printf(“\%d\n”, scanf(“\%d”, &i)); The scanf function returns the value 1(one). Therefore, the output of the program is ‘1’.

What will the output int main () int VAR 010 printf \%d var );?

Since gPtr isn’t initialized in the program, it’ll print sometimes Equal and at other times Not Equal. Question 10 Explanation: It should be noted that global variables such gPtr (which is a global pointer to int) are initialized to ZERO….Discuss it.

A	Compiler Error: var is not defined
C	0

What is the output of this program #include?

5. What is the output of this program? Explanation: The output is garbage value.

What is an output statement?

An input/output statement or IO statement is a portion of a program that instructs a computer how to read and process information. It pertains to gather information from an input device, or sending information to an output device.

How does printf work in C?

In your particular Example, printf evaluates a++ first, reads 10 and prints it and only then increments it to 11. printf then evaluates ++a, increments it first, reads 12 and prints it out. The last variable printf evaluates is read as it is (12) and is printed without any change.

What does \%D mean in printf?

printf(\%d) is a syntax error. \% is the remainder operator. It requires two operands, and you’ve only given it one (assuming d is a valid expression). I’m sure that’s not what you meant, but it’s what you wrote.

Is it possible to express a 128 bit integer in GCC?

There is no support in GCC to express an integer constant of type __int128 for targets having long long integer with less then [ sic ] 128 bit width. Interestingly, although that does not mention __uint128_t, that type is accepted, even with stringent warnings set:

What is uint128_t in GCC?

Note that uint128_t is my own typedef for __uint128_t. The GCC 4.7.1 manual says: As an extension the integer scalar type __int128 is supported for targets having an integer mode wide enough to hold 128-bit.