Online C Programming Test - C Programming Test 6
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- Total number of questions: 20.
- Time allotted: 20 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.
#include<stdio.h>
int main()
{
short int i = 0;
for(i<=5 && i>=-1; ++i; i>0)
printf("%u,", i);
return 0;
}
for(i<=5 && i>=-1; ++i; i>0) so expression i<=5 && i>=-1 initializes for loop. expression ++i is the loop condition. expression i>0 is the increment expression.
In for( i <= 5 && i >= -1; ++i; i>0) expression i<=5 && i>=-1 evaluates to one.
Loop condition always get evaluated to true. Also at this point it increases i by one.
An increment_expression i>0 has no effect on value of i.so for loop get executed till the limit of integer (ie. 65535)
#include<stdio.h>
int main()
{
float a = 0.7;
if(0.7 > a)
printf("Hi\n");
else
printf("Hello\n");
return 0;
}
if(0.7 > a) here a is a float variable and 0.7 is a double constant. The double constant 0.7 is greater than the float variable a. Hence the if condition is satisfied and it prints 'Hi'
Example:
#include<stdio.h>
int main()
{
float a=0.7;
printf("%.10f %.10f\n",0.7, a);
return 0;
}
Output:
0.7000000000 0.6999999881
| 1: | Every if-else statement can be replaced by an equivalent statements using ?: operators |
| 2: | Nested if-else statements are allowed. |
| 3: | Multiple statements in an if block are allowed. |
| 4: | Multiple statements in an else block are allowed. |
#include<stdio.h>
#include<math.h>
int main()
{
printf("%f\n", sqrt(36.0));
return 0;
}
printf("%f\n", sqrt(36.0)); It prints the square root of 36 in the float format(i.e 6.000000).
Declaration Syntax: double sqrt(double x) calculates and return the positive square root of the given number.
Yes, too many recursive calls may result into stack overflow. because when a function is called its return address is stored in stack.
After sometime the stack memory will be filled completely. Hence stack overflow error will occur.
FALSE, The macro is case insensitive.
False, Always the macro is substituted by the given text/expression.
True, the header file contains classes, function prototypes, structure declaration, macros.
Declaration:
char *strnset(char *s, int ch, size_t n); Sets the first n characters of s to ch
#include <stdio.h>
#include <string.h>
int main(void)
{
char *string = "abcdefghijklmnopqrstuvwxyz";
char letter = 'x';
printf("string before strnset: %s\n", string);
strnset(string, letter, 13);
printf("string after strnset: %s\n", string);
return 0;
}
Output:
string before strnset: abcdefghijklmnopqrstuvwxyz
string after strnset: xxxxxxxxxxxxxnopqrstuvwxyz
#include<stdio.h>
int main()
{
char str = "IndiaBIX";
printf("%s\n", str);
return 0;
}
The line char str = "IndiaBIX"; generates "Non portable pointer conversion" error.
To eliminate the error, we have to change the above line to
char *str = "IndiaBIX"; (or) char str[] = "IndiaBIX";
Then it prints "IndiaBIX".
#include<stdio.h>
int main()
{
char str1[] = "Hello";
char str2[10];
char *t, *s;
s = str1;
t = str2;
while(*t=*s)
*t++ = *s++;
printf("%s\n", str2);
return 0;
}
#include<stdio.h>
int main()
{
char *str[] = {"Frogs", "Do", "Not", "Die", "They", "Croak!"};
printf("%d, %d", sizeof(str), strlen(str[0]));
return 0;
}
Step 1: char *str[] = {"Frogs", "Do", "Not", "Die", "They", "Croak!"}; The variable str is declared as an pointer to the array of 6 strings.
Step 2: printf("%d, %d", sizeof(str), strlen(str[0]));
sizeof(str) denotes 6 * 4 bytes = 24 bytes. Hence it prints '24'
strlen(str[0])); becomes strlen(Frogs)). Hence it prints '5';
Hence the output of the program is 24, 5
Hint: If you run the above code in 16 bit platform (Turbo C under DOS) the output will be 12, 5. Because the pointer occupies only 2 bytes. If you run the above code in Linux (32 bit platform), the output will be 24, 5 (because the size of pointer is 4 bytes).
#include<stdio.h>
int main()
{
FILE *fp;
fp=fopen("trial", "r");
fseek(fp, "20", SEEK_SET);
fclose(fp);
return 0;
}
cmd> myprog friday tuesday sunday
/* myprog.c */
#include<stdio.h>
int main(int argc, char *argv[])
{
printf("%c", *++argv[1]);
return 0;
}
#include<stdio.h>
int main()
{
const int x=5;
const int *ptrx;
ptrx = &x;
*ptrx = 10;
printf("%d\n", x);
return 0;
}
Step 1: const int x=5; The constant variable x is declared as an integer data type and initialized with value '5'.
Step 2: const int *ptrx; The constant variable ptrx is declared as an integer pointer.
Step 3: ptrx = &x; The address of the constant variable x is assigned to integer pointer variable ptrx.
Step 4: *ptrx = 10; Here we are indirectly trying to change the value of the constant vaiable x. This will result in an error.
To change the value of const variable x we have to use *(int *)&x = 10;
#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.
#include<stdio.h>
const char *fun();
int main()
{
char *ptr = fun();
return 0;
}
const char *fun()
{
return "Hello";
}
#include<stdio.h>
int main()
{
int i;
char c;
for(i=1; i<=5; i++)
{
scanf("%c", &c); /* given input is 'a' */
printf("%c", c);
ungetc(c, stdin);
}
return 0;
}
for(i=1; i<=5; i++) Here the for loop runs 5 times.
Loop 1:
scanf("%c", &c); Here we give 'a' as input.
printf("%c", c); prints the character 'a' which is given in the previous "scanf()" statement.
ungetc(c, stdin); "ungetc()" function pushes character 'a' back into input stream.
Loop 2:
Here the scanf("%c", &c); get the input from "stdin" because of "ungetc" function.
printf("%c", c); Now variable c = 'a'. So it prints the character 'a'.
ungetc(c, stdin); "ungetc()" function pushes character 'a' back into input stream.
This above process will be repeated in Loop 3, Loop 4, Loop 5.