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
Question
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
Solution
The output of this program will be "sizeof(i) = 1". This is because the '#define' preprocessor directive is used to replace 'int' with 'char' everywhere in the code. Since the size of a 'char' is 1 byte, the 'sizeof(i)' will return 1.
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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
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