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- Question
What will be the output of the program?
#include<stdio.h> int main() { char str1[] = "Hello"; char str2[] = "Hello"; if(str1 == str2) printf("Equal\n"); else printf("Unequal\n"); return 0; }
Options- A. Equal
- B. Unequal
- C. Error
- D. None of above
- Correct Answer
- Unequal
ExplanationStep 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".
More questions
- 1. What will be the output of the program?
#include<stdio.h> int main() { static char *s[] = {"black", "white", "pink", "violet"}; char **ptr[] = {s+3, s+2, s+1, s}, ***p; p = ptr; ++p; printf("%s", **p+1); return 0; }
Options- A. ink
- B. ack
- C. ite
- D. let Discuss
Correct Answer: ink
- 2. What will be the output of the program?
#include<stdio.h> int main() { int arr[3] = {2, 3, 4}; char *p; p = arr; p = (char*)((int*)(p)); printf("%d, ", *p); p = (int*)(p+1); printf("%d", *p); return 0; }
Options- A. 2, 3
- B. 2, 0
- C. 2, Garbage value
- D. 0, 0 Discuss
Correct Answer: 2, 0
- 3. Will it result in to an error if a header file is included twice?
Options- A. Yes
- B. No
- C. It is compiler dependent. Discuss
Correct Answer: It is compiler dependent.
Explanation:
Unless the header file has taken care to ensure that if already included it doesn't get included again.
Turbo C, GCC compilers would take care of these problems, generate no error.- 4. While defining a variable argument list function we drop the ellipsis(...)?
Options- A. Yes
- B. No Discuss
Correct Answer: Yes
- 5. It is not possible to create an array of pointer to structures.
Options- A. True
- B. False Discuss
Correct Answer: False
- 6.
1 : typedef long a;
extern int a c;2 : typedef long a;
extern a int c;3 : typedef long a;
extern a c;
Options- A. 1 correct
- B. 2 correct
- C. 3 correct
- D. 1, 2, 3 are correct Discuss
Correct Answer: 3 correct
Explanation:
typedef long a;
extern int a c; while compiling this statement becomes extern int long c;. This will result in to "Declaration syntax error".typedef long a;
extern a int c; while compiling this statement becomes extern long int c;. This will result in to "Too many types in declaration error".typedef long a;
extern a c; while compiling this statement becomes extern long c;. This is a valid c declaration statement. It says variable c is long data type and defined in some other file or module.So, Option C is the correct answer.
- 7. Which of the declaration is correct?
Options- A. int length;
- B. char int;
- C. int long;
- D. float double; Discuss
Correct Answer: int length;
Explanation:
int length; denotes that variable length is int(integer) data type.char int; here int is a keyword cannot be used a variable name.
int long; here long is a keyword cannot be used a variable name.
float double; here double is a keyword cannot be used a variable name.
So, the answer is int length;(Option A).
- 8. What will be the output of the program?
#include<stdio.h> int main() { int i=2; int j = i + (1, 2, 3, 4, 5); printf("%d\n", j); return 0; }
Options- A. 4
- B. 7
- C. 6
- D. 5 Discuss
Correct Answer: 7
Explanation:
Because, comma operator used in the expression i (1, 2, 3, 4, 5). The comma operator has left-right associativity. The left operand is always evaluated first, and the result of evaluation is discarded before the right operand is evaluated. In this expression 5 is the right most operand, hence after evaluating expression (1, 2, 3, 4, 5) the result is 5, which on adding to i results into 7.- 9. What will be the output of the program?
#include<stdio.h> int main() { int i=-3, j=2, k=0, m; m = ++i && ++j && ++k; printf("%d, %d, %d, %d\n", i, j, k, m); return 0; }
Options- A. -2, 3, 1, 1
- B. 2, 3, 1, 2
- C. 1, 2, 3, 1
- D. 3, 3, 1, 2 Discuss
Correct Answer: -2, 3, 1, 1
Explanation:
Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.Step 2: m = ++i && ++j && ++k;
becomes m = -2 && 3 && 1;
becomes m = TRUE && TRUE; Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.
Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of i,j,k are increemented by '1'(one).
Hence the output is "-2, 3, 1, 1".
- 10. What will be the output of the program?
#include<stdio.h> int i; int fun1(int); int fun2(int); int main() { extern int j; int i=3; fun1(i); printf("%d,", i); fun2(i); printf("%d", i); return 0; } int fun1(int j) { printf("%d,", ++j); return 0; } int fun2(int i) { printf("%d,", ++i); return 0; } int j=1;
Options- A. 3, 4, 4, 3
- B. 4, 3, 4, 3
- C. 3, 3, 4, 4
- D. 3, 4, 3, 4 Discuss
Correct Answer: 4, 3, 4, 3
Explanation:
Step 1: int i; The variable i is declared as an global and integer type.Step 2: int fun1(int); This prototype tells the compiler that the fun1() accepts the one integer parameter and returns the integer value.
Step 3: int fun2(int); This prototype tells the compiler that the fun2() accepts the one integer parameter and returns the integer value.
Step 4: extern int j; Inside the main function, the extern variable j is declared and defined in another source file.
Step 5: int i=3; The local variable i is defines as an integer type and initialized to 3.
Step 6: fun1(i); The fun1(i) increements the given value of variable i prints it. Here fun1(i) becomes fun1(3) hence it prints '4' then the control is given back to the main function.
Step 7: printf("%d,", i); It prints the value of local variable i. So, it prints '3'.
Step 8: fun2(i); The fun2(i) increements the given value of variable i prints it. Here fun2(i) becomes fun2(3) hence it prints '4' then the control is given back to the main function.
Step 9: printf("%d,", i); It prints the value of local variable i. So, it prints '3'.
Hence the output is "4 3 4 3".
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- 1. What will be the output of the program?