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- Question
What will function gcvt() do?
Options- A. Convert vector to integer value
- B. Convert floating-point number to a string
- C. Convert 2D array in to 1D array.
- D. Covert multi Dimensional array to 1D array
- Correct Answer
- Convert floating-point number to a string
ExplanationThe gcvt() function converts a floating-point number to a string. It converts given value to a null-terminated string.#include <stdlib.h> #include <stdio.h> int main(void) { char str[25]; double num; int sig = 5; /* significant digits */ /* a regular number */ num = 9.876; gcvt(num, sig, str); printf("string = %s\n", str); /* a negative number */ num = -123.4567; gcvt(num, sig, str); printf("string = %s\n", str); /* scientific notation */ num = 0.678e5; gcvt(num, sig, str); printf("string = %s\n", str); return(0); }Output:
string = 9.876
string = -123.46
string = 67800
Library Functions problems
Search Results
- 1. What will be the output of the program?
#include<stdio.h> #include<math.h> int main() { float i = 2.5; printf("%f, %d", floor(i), ceil(i)); return 0; }
Options- A. 2, 3
- B. 2.000000, 3
- C. 2.000000, 0
- D. 2, 0 Discuss
Correct Answer: 2.000000, 0
Explanation:
Both ceil() and floor() return the integer found as a double.floor(2.5) returns the largest integral value(round down) that is not greater than 2.5. So output is 2.000000.
ceil(2.5) returns 3, while converting the double to int it returns '0'.
So, the output is '2.000000, 0'.- 2. What will be the output of the program?
#include<stdio.h> int main() { int i; i = scanf("%d %d", &i, &i); printf("%d\n", i); return 0; }
Options- A. 1
- B. 2
- C. Garbage value
- D. Error: cannot assign scanf to variable Discuss
Correct Answer: 2
Explanation:
scanf() returns the number of variables to which you are provding the input. i = scanf("%d %d", &i, &i); Here Scanf() returns 2. So i = 2.printf("%d\n", i); Here it prints 2.
- 3. What will be the output of the program?
#include<stdio.h> #include<string.h> int main() { char dest[] = {97, 97, 0}; char src[] = "aaa"; int i; if((i = memcmp(dest, src, 2))==0) printf("Got it"); else printf("Missed"); return 0; }
Options- A. Missed
- B. Got it
- C. Error in memcmp statement
- D. None of above Discuss
Correct Answer: Got it
Explanation:
memcmp compares the first 2 bytes of the blocks dest and src as unsigned chars. So, the ASCII value of 97 is 'a'.
if((i = memcmp(dest, src, 2))==0) When comparing the array dest and src as unsigned chars, the first 2 bytes are same in both variables.so memcmp returns '0'.
Then, the if(0=0) condition is satisfied. Hence the output is "Got it".- 4. The macro va_arg is used to extract an argument from the fixed micro argument list and advance the pointer to the next argument.
Options- A. Yes
- B. No Discuss
Correct Answer: No
- 5. Is it necessary that in a function which accepts variable argument list there should be at least be one fixed argument?
Options- A. Yes
- B. No Discuss
Correct Answer: Yes
- 6. What will be the output of the program?
#include<stdio.h> int main() { int i; i = printf("How r u\n"); i = printf("%d\n", i); printf("%d\n", i); return 0; }
Options- A. How r u
7
2 - B. How r u
8
2 - C. How r u
1
1 - D. Error: cannot assign printf to variable Discuss
Correct Answer: How r u
8
2
Explanation:
In the program, printf() returns the number of charecters printed on the console i = printf("How r u\n"); This line prints "How r u" with a new line character and returns the length of string printed then assign it to variable i.
So i = 8 (length of '\n' is 1).i = printf("%d\n", i); In the previous step the value of i is 8. So it prints "8" with a new line character and returns the length of string printed then assign it to variable i. So i = 2 (length of '\n' is 1).
printf("%d\n", i); In the previous step the value of i is 2. So it prints "2".
- 7. What will be the output of the program?
#include<stdio.h> int main() { int i; char c; for(i=1; i<=5; i++) { scanf("%c", &c); /* given input is 'b' */ ungetc(c, stdout); printf("%c", c); ungetc(c, stdin); } return 0; }
Options- A. bbbb
- B. bbbbb
- C. b
- D. Error in ungetc statement. Discuss
Correct Answer: b
Explanation:
The ungetc() function pushes the character c back onto the named input stream, which must be open for reading.This character will be returned on the next call to getc or fread for that stream.
One character can be pushed back in all situations.
A second call to ungetc without a call to getc will force the previous character to be forgotten.
- 8. What will be the output of the program?
#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; }
Options- A. aaaa
- B. aaaaa
- C. Garbage value.
- D. Error in ungetc statement. Discuss
Correct Answer: aaaaa
Explanation:
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.
- 9. What will be the output of the program?
#include<stdio.h> #include<stdlib.h> int main() { char *i = "55.555"; int result1 = 10; float result2 = 11.111; result1 = result1+atoi(i); result2 = result2+atof(i); printf("%d, %f", result1, result2); return 0; }
Options- A. 55, 55.555
- B. 66, 66.666600
- C. 65, 66.666000
- D. 55, 55 Discuss
Correct Answer: 65, 66.666000
Explanation:
Function atoi() converts the string to integer.
Function atof() converts the string to float.result1 = result1+atoi(i);
Here result1 = 10 + atoi(55.555);
result1 = 10 + 55;
result1 = 65;result2 = result2+atof(i);
Here result2 = 11.111 + atof(55.555);
result2 = 11.111 + 55.555000;
result2 = 66.666000;
So the output is "65, 66.666000" .- 10. What will be the output of the program?
#include<stdio.h> int main() { int x, y, z; x=y=z=1; z = ++x || ++y && ++z; printf("x=%d, y=%d, z=%d\n", x, y, z); return 0; }
Options- A. x=2, y=1, z=1
- B. x=2, y=2, z=1
- C. x=2, y=2, z=2
- D. x=1, y=2, z=1 Discuss
Correct Answer: x=2, y=1, z=1
Explanation:
Step 1: x=y=z=1; here the variables x ,y, z are initialized to value '1'.Step 2: z = ++x || ++y && ++z; becomes z = ( (++x) || (++y && ++z) ). Here ++x becomes 2. So there is no need to check the other side because ||(Logical OR) condition is satisfied.(z = (2 || ++y && ++z)). There is no need to process ++y && ++z. Hence it returns '1'. So the value of variable z is '1'
Step 3: printf("x=%d, y=%d, z=%d\n", x, y, z); It prints "x=2, y=1, z=1". here x is increemented in previous step. y and z are not increemented.
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- 1. What will be the output of the program?