1. | What will be the output of the program? #include<stdio.h> int main() |
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Answer: Option B Explanation: Step 1: const c = -11; The constant variable 'c' is declared and initialized to value "-11". Step 2: const int d = 34; The constant variable 'd' is declared as an integer and initialized to value '34'. Step 3: printf("%d, %d\n", c, d); The value of the variable 'c' and 'd' are printed. Hence the output of the program is -11, 34 |
2. | What will be the output of the program? #include<stdio.h> int main() |
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Answer: Option D Explanation: This program will show an error "Cannot modify a const object". Step 1: const int i=0; The constant variable 'i' is declared as an integer and initialized with value of '0'(zero). Step 2: printf("%d\n", i++); Here the variable 'i' is increemented by 1(one). This will create an error "Cannot modify a const object". Because, we cannot modify a const variable. |
3. | What will be the output of the program (in Turbo C)? #include<stdio.h> int fun(int *f) |
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Answer: Option A Explanation: Step 1: const int arr[5] = {1, 2, 3, 4, 5}; The constant variable arr is declared as an integer array and initialized to arr[0] = 1, arr[1] = 2, arr[2] = 3, arr[3] = 4, arr[4] = 5 Step 2: printf("Before modification arr[3] = %d", arr[3]); It prints the value of arr[3] (ie. 4). Step 3: fun(&arr[3]); The memory location of the arr[3] is passed to fun() and arr[3] value is modified to 10. A const variable can be indirectly modified by a pointer. Step 4: printf("After modification arr[3] = %d", arr[3]); It prints the value of arr[3] (ie. 10). Hence the output of the program is Before modification arr[3] = 4 After modification arr[3] = 10 |
4. | What will be the output of the program? #include<stdio.h> int main() |
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Answer: Option C Explanation: Step 1: int get(); This is the function prototype for the funtion get(), it tells the compiler returns an integer value and accept no parameters. Step 2: const int x = get(); The constant variable x is declared as an integer data type and initialized with the value "20". The function get() returns the value "20". Step 3: printf("%d", x); It prints the value of the variable x. Hence the output of the program is "20". |
5. | What will be the output of the program? #include<stdio.h> int main() return 0; |
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Answer: Option C Explanation: Step 1: const char *s = ""; The constant variable s is declared as an pointer to an array of characters type and initialized with an empty string. Step 2: char str[] = "Hello"; The variable str is declared as an array of charactrers type and initialized with a string "Hello". Step 3: s = str; The value of the variable str is assigned to the variable s. Therefore str contains the text "Hello". Step 4: while(*s){ printf("%c", *s++); } Here the while loop got executed untill the value of the variable s is available and it prints the each character of the variable s. Hence the output of the program is "Hello". |
6. | What will be the output of the program in TurboC? #include<stdio.h> int main() |
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Answer: Option B Explanation: No answer description available for this question. Let us discuss.
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7. | What will be the output of the program? #include<stdio.h> int main() |
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Answer: Option C Explanation: 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; |
8. | What will be the output of the program? #include<stdio.h> int main() |
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Answer: Option C Explanation: No answer description available for this question. Let us discuss.
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