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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;}

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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;}

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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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