Online C Programming Test - C Programming Test - Random
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- Total number of questions: 20.
- Time allotted: 30 minutes.
- Each question carries 1 mark; there are no negative marks.
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- All the best!
Marks : 2/20
Test Review : View answers and explanation for this test.
A function prototype in C or C++ is a declaration of a function that omits the function body but does specify the function's name, argument types and return type.
While a function definition specifies what a function does, a function prototype can be thought of as specifying its interface.
True, we can use long double; if double range is not enough.
double = 8 bytes.
long double = 10 bytes.
#include<stdio.h>
int main()
{
int j=1;
while(j <= 255)
{
printf("%c %d\n", j, j);
j++;
}
return 0;
}
The while(j <= 255) loop will get executed 255 times. The size short int(2 byte wide) does not affect the while() loop.
#include<stdio.h>
int main()
{
struct emp
{
char name[20];
float sal;
};
struct emp e[10];
int i;
for(i=0; i<=9; i++)
scanf("%s %f", e[i].name, &e[i].sal);
return 0;
}
Compile and Run the above program in Turbo C:
C:\>myprogram.exe Sundar 2555.50 scanf : floating point formats not linked Abnormal program termination
The program terminates abnormally at the time of entering the float value for e[i].sal.
Solution:
Just add the following function in your program. It will force the compiler to include required libraries for handling floating point linkages.
static void force_fpf() /* A dummy function */
{
float x, *y; /* Just declares two variables */
y = &x; /* Forces linkage of FP formats */
x = *y; /* Suppress warning message about x */
}True, The default return type for a function is int.
#include<stdio.h>
#define CUBE(x) (x*x*x)
int main()
{
int a, b=3;
a = CUBE(b++);
printf("%d, %d\n", a, b);
return 0;
}
The macro function CUBE(x) (x*x*x) calculates the cubic value of given number(Eg: 103.)
Step 1: int a, b=3; The variable a and b are declared as an integer type and varaible b id initialized to 3.
Step 2: a = CUBE(b++); becomes
=> a = b++ * b++ * b++;
=> a = 3 * 3 * 3; Here we are using post-increement operator, so the 3 is not incremented in this statement.
=> a = 27; Here, 27 is store in the variable a. By the way, the value of variable b is incremented by 3. (ie: b=6)
Step 3: printf("%d, %d\n", a, b); It prints the value of variable a and b.
Hence the output of the program is 27, 6.
#include<stdio.h>
int main()
{
char *str;
str = "%s";
printf(str, "K\n");
return 0;
}
#include<stdio.h>
int main()
{
char str1[] = "Hello";
char str2[] = "Hello";
if(str1 == str2)
printf("Equal\n");
else
printf("Unequal\n");
return 0;
}
Step 1: char str1[] = "Hello"; The variable str1 is declared as an array of characters and initialized with a string "Hello".
Step 2: char str2[] = "Hello"; The variable str2 is declared as an array of characters and initialized with a string "Hello".
We have use strcmp(s1,s2) function to compare strings.
Step 3: if(str1 == str2) here the address of str1 and str2 are compared. The address of both variable is not same. Hence the if condition is failed.
Step 4: At the else part it prints "Unequal".
#include<stdio.h>
int main()
{
char a[] = "Visual C++";
char *b = "Visual C++";
printf("%d, %d\n", sizeof(a), sizeof(b));
printf("%d, %d", sizeof(*a), sizeof(*b));
return 0;
}
#include<stdio.h>
int main()
{
int a=250;
printf("%1d \n", a);
return 0;
}
int a=250; The variable a is declared as an integer type and initialized to value 250.
printf("%1d \n", a); It prints the value of variable a.
Hence the output of the program is 250.
True, we should not be able to read a file after writing in that file without calling the below functions.
int fflush ( FILE * stream ); If the given stream was open for writing and the last i/o operation was an output operation, any unwritten data in the output buffer is written to the file.
int fseek ( FILE * stream, long int offset, int origin ); Its purpose is to change the file position indicator for the specified stream.
void rewind ( FILE * stream ); Sets the position indicator associated with stream to the beginning of the file.
cmd> myprog one two three
/* myprog.c */
#include<stdio.h>
int main(int argc, char *argv[])
{
int i;
for(i=1; i<argc; i++)
printf("%c", argv[i][0]);
return 0;
}
/* myproc.c */
#include<stdio.h>
int main(int argc, char *argv[])
{
printf("%s", argv[0]);
return 0;
}
#include<stdio.h>
#define MAX 128
int main()
{
const int max=128;
char array[max];
char string[MAX];
array[0] = string[0] = 'A';
printf("%c %c\n", array[0], string[0]);
return 0;
}
Step 1: A macro named MAX is defined with value 128
Step 2: const int max=128; The constant variable max is declared as an integer data type and it is initialized with value 128.
Step 3: char array[max]; This statement reports an error "constant expression required". Because, we cannot use variable to define the size of array.
To avoid this error, we have to declare the size of an array as static. Eg. char array[10]; or use macro char array[MAX];
Note: The above program will print A A as output in Unix platform.
int *f();char *scr;#include<stdio.h>
int main()
{
char str[] = "IndiaBIX";
printf("%.#s %2s", str, str);
return 0;
}