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- Question
What will be the output of the program?
#include<stdio.h> int main() { char str[25] = "CuriousTab"; printf("%s\n", &str+2); return 0; }
Options- A. Garbage value
- B. Error
- C. No output
- D. diaCURIOUSTAB
- Correct Answer
- Garbage value
ExplanationStep 1: char str[25] = "CuriousTab"; The variable str is declared as an array of characteres and initialized with a string "CuriousTab".Step 2: printf("%s\n", &str+2);
=> In the printf statement %s is string format specifier tells the compiler to print the string in the memory of &str+2
=> &str is a location of string "CuriousTab". Therefore &str+2 is another memory location.
Hence it prints the Garbage value.
More questions
- 1. It is necessary to call the macro va_end if va_start is called in the function.
Options- A. Yes
- B. No Discuss
Correct Answer: Yes
- 2. Left shifting a number by 1 is always equivalent to multiplying it by 2.
Options- A. True
- B. False Discuss
Correct Answer: True
Explanation:
0001 => 1
0010 => 2
0100 => 4
1000 => 8
- 3. What will be the output of the program?
#include<stdio.h> int main() { printf("%%%%\n"); return 0; }
Options- A. %%%%%
- B. %%
- C. No output
- D. Error Discuss
Correct Answer: %%
- 4. A function that receives variable number of arguments should use va_arg() to extract arguments from the variable argument list.
Options- A. True
- B. False Discuss
Correct Answer: True
- 5. What will be the output of the program?
#include<stdio.h> int main() { float a = 0.7; if(0.7 > a) printf("Hi\n"); else printf("Hello\n"); return 0; }
Options- A. Hi
- B. Hello
- C. Hi Hello
- D. None of above Discuss
Correct Answer: Hi
Explanation:
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- 6. What will be the output of the program (sample.c) given below if it is executed from the command line?
cmd> sample friday tuesday sunday/* sample.c */ #include<stdio.h> int main(int sizeofargv, char *argv[]) { while(sizeofargv) printf("%s", argv[--sizeofargv]); return 0; }
Options- A. sample friday tuesday sunday
- B. sample friday tuesday
- C. sunday tuesday friday sample
- D. sunday tuesday friday Discuss
Correct Answer: sunday tuesday friday sample
- 7. What will be the output of the program (sample.c) given below if it is executed from the command line?
cmd> sample friday tuesday sunday/* sample.c */ #include<stdio.h> int main(int argc, char *argv[]) { printf("%c", *++argv[2] ); return 0; }
Options- A. s
- B. f
- C. u
- D. r Discuss
Correct Answer: u
- 8. Point out the error in the program.
#include<stdio.h> int main() { const int k=7; int *const q=&k; printf("%d", *q); return 0; }
Options- A. Error: RValue required
- B. Error: Lvalue required
- C. Error: cannot convert from 'const int *' to 'int *const'
- D. No error Discuss
Correct Answer: No error
Explanation:
No error. This will produce 7 as output.- 9. Bit fields CANNOT be used in union.
Options- A. True
- B. False Discuss
Correct Answer: False
Explanation:
The following is the example program to explain "using bit fields inside an union".#include<stdio.h> union Point { unsigned int x:4; unsigned int y:4; int res; }; int main() { union Point pt; pt.x = 2; pt.y = 3; pt.res = pt.y; printf("\n The value of res = %d" , pt.res); return 0; } // Output: The value of res = 3- 10. What will be the output of the program (myprog.c) given below if it is executed from the command line?
cmd> myprog 10 20 30/* myprog.c */ #include<stdio.h> int main(int argc, char **argv) { int i; for(i=0; i<argc; i++) printf("%s\n", argv[i]); return 0; }
Options- A. 10 20 30
- B. myprog 10 20
- C. myprog 10 20 30
- D. 10 20 Discuss
Correct Answer: myprog 10 20 30
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