MCQs
float a=3.15529; The variable a is declared as an float data type and initialized
to value 3.15529;
printf("%2.1f`setminus`n", a); The precision specifier tells .1f tells the printf function
to place only one number after the .(dot).
Hence the output is 3.2
No answer description available for this question.
float a=3.15529; The variable a is declared as an float data type and initialized to value 3.15529;
printf("%2.1fn", a); The precision specifier tells .1f tells the printf function to place only one number
after the .(dot).
Hence the output is 3.2
If the file 'source.txt' contains a line "Be my friend" which of the following will
be the output of below program?
#include<stdio.h>
int main()
{
FILE *fs, *ft;
char c[10];
fs = fopen("source.txt", "r");
c[0] = getc(fs);
fseek(fs, 0, SEEK_END);
fseek(fs, -3L, SEEK_CUR);
fgets(c, 5, fs);
puts(c);
return 0;
}
The file source.txt contains "Be my friend".
fseek(fs, 0, SEEK_END); moves the file pointer to the end of the file.
fseek(fs, -3L, SEEK_CUR); moves the file pointer backward by 3 characters.
fgets(c, 5, fs); read the file from the current position of the file pointer.
Hence, it contains the last 3 characters of "Be my friend".
Therefore, it prints "end".
Step 1: int k=1; The variable k is declared as an integer type and initialized to '1'.
Step 2: printf("%d == 1 is" "%sn", k, k==1?"TRUE":"FALSE"); becomes
=> k==1?"TRUE":"FALSE"
=> 1==1?"TRUE":"FALSE"
=> "TRUE"
Therefore the output of the program is 1 == 1 is TRUE
Here fputc('A', fp1); stores 'A' in the file1.c then fputc('B', fp2); overwrites the contents of the file1.c
with value 'B'. Because the fp1 and fp2 opens the file1.c in write mode.
Hence the file1.c contents is 'B'.
To scan a float value, %f is used as format specifier.
To scan a double value, %lf is used as format specifier.
Therefore, the answer is scanf("%f %lf", &a, &b);
Because, In fgets() we can specify the size of the buffer into which the string supplied will be stored.