[Solved] Point out the error, if any in the program. #include<stdio.h> int main() { int i = 1; switch(i) { printf('This is c program.'); case 1: printf('Case1'); break; case 2: printf('Case2'); break; } return 0; } - Programming - C Programming - Control Instructions

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Q1: Point out the error, if any in the program.

#include<stdio.h>
int main()
{
    int i = 1;
    switch(i)
    {
        printf('This is c program.');
        case 1:
            printf('Case1');
            break;
        case 2:
            printf('Case2');
            break;
    }
return 0;
}

A Error: No default specified

Error: Invalid printf statement after switch statement

C No Error and prints "Case1"

D None of above

ANS:A - Error: No default specified

switch(i) becomes switch(1), then the case 1: block is get executed. Hence it prints 'Case1'. printf('This is c program.'); is ignored by the compiler. Hence there is no error and prints 'Case1'.

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Q1: Point out the error, if any in the program.
`#include<stdio.h> int main() { int i = 1; switch(i) { printf('This is c program.'); case 1: printf('Case1'); break; case 2: printf('Case2'); break; } return 0; }`

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Q1: Point out the error, if any in the program. #include<stdio.h> int main() { int i = 1; switch(i) { printf('This is c program.'); case 1: printf('Case1'); break; case 2: printf('Case2'); break; } return 0; }

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Q1: Point out the error, if any in the program.
`#include<stdio.h> int main() { int i = 1; switch(i) { printf('This is c program.'); case 1: printf('Case1'); break; case 2: printf('Case2'); break; } return 0; }`

For help Students Orientation
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