lesWhat will be the status of this program upon compilation? #include <stdio.h>int main() { int i; i = 10; int I = 20; printf("Values i = %d and I = %d\n", i, I); getchar(); return 0;}
Question
lesWhat will be the status of this program upon compilation? #include <stdio.h>int main() { int i; i = 10; int I = 20; printf("Values i = %d and I = %d\n", i, I); getchar(); return 0;}
Solution
The program will compile successfully. The variables 'i' and 'I' are treated as two different variables because C is a case-sensitive language. Therefore, the program will print "Values i = 10 and I = 20" as output.
Similar Questions
lesWhat will be the status of this program upon compilation? #include <stdio.h>int main() { int i; i = 10; int I = 20; printf("Values i = %d and I = %d\n", i, I); getchar(); return 0;}
ablesWhat will be the status of this program upon compilation?#include <stdio.h>int main() { char j = 'a'; int i = j; int k = (int)j; int l = 75; char m = (char)l; printf("value of i = %d, k = %d\n", i, k); printf("value of l = %d, m = %c\n", l, m); getchar(); return 0;}
What will be the output of this program? (on a standard 64 bits, Linux machine)#include <stdio.h>#include <stdlib.h>#define int charint main(void){ int i; i = 5; printf ("sizeof(i) = %lu", sizeof(i)); return (EXIT_SUCCESS);}It does not compileSegmentation Faultsizeof(i) = 4sizeof(i) = 1sizeof(i) = 8sizeof(i) = 5
ablesWhat will be the status of this program upon compilation?#include <stdio.h>/* integer variable program */int main() { int i; i = 10; printf("Value of i is %d\n", i); i = 20; printf("Value of i is %d\n", i); getchar(); return 0;}OptionsError in compilationWill run successfully and display the following,Value of i is 10Value of i is 20Will run successfully and display the following,Value of i is 10Value of i is 10Garbage Values
What happens if we compile the following code?
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