What will be output of the below C code in 32-bit machine..?

Question

 #include<stdio.h>
    double i;
    int main()
    {
    (int)(float)(char) i;
    printf("%d",sizeof(i));
    return 0;
    }

Its showing output as 8. Can anyone please explain me why it is showing as 8.? Does the Typecasting have any effect on the Variable i..so that the output can possibly be 4??

`(int)(float)(char) i` that does nothing. `sizeof(double)` is apparently 8 on your machine, which isn't surprising. — Jean-François Fabre, Sep 04 '19 at 13:32
4 bytes (32-bits) is too small for a `double` which is usually 64 bits. IIRC, needs at least 39 bits. [Any guaranteed minimum sizes for types in C?](https://stackoverflow.com/q/1738568/2410359) — chux - Reinstate Monica, Sep 04 '19 at 13:33
[The proper format specifier for `size_t` is `%zu`](https://stackoverflow.com/questions/2524611/how-can-one-print-a-size-t-variable-portably-using-the-printf-family). `z` is the length modifier for `size_t`, and `u` as `size_t` is an unsigned value. You could also use `%zx` for hex output or `%zo` for octal output. — Andrew Henle, Sep 04 '19 at 13:42
You made a change. The answer specifically says that the size of *a double cast to an int* is 4 in that environment. You didn't get the size of a double cast to an int (and you may not be using the same environment) — ikegami, Sep 04 '19 at 13:49

score 1 · Accepted Answer · answered Sep 04 '19 at 14:37

1

(int)(float)(char) i; is not a definition of i. It merely is using the value for nothing.

#include <stdio.h>
double i;
int main(void) {
    i; // use i for nothing
    (int)i; // convert the value of i to integer, than use that value for nothing
    (int)(float)i; // convert to float, then to int, then use for nothing
    (int)(float)(char)i; // convert char, then to float, then to int, then use for nothing
    printf("sizeof i is %d\n", (int)sizeof i);
    char i; // define a new i (and hide the previous one) of type char
    printf("sizeof i is %d\n", (int)sizeof i);
}

answered Sep 04 '19 at 14:37

pmg

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printf("sizeof i is %d\n", (int)sizeof i); Can you explain what does (int)sizeof i does in this statement. It is showing the Output value as 8 And what is the difference between (int)sizeof i and sizeof((int) i) – Kannan Mrithunjai Sep 04 '19 at 15:05
`sizeof i` yelds the size of the type of `i`; `(int)sizeof i` converts the size of type of `i` (type `size_t`) to integer so that it can be printed safely with `"%d"` in `printf()`. `sizeof(int)` (or `sizeof((int)i)`) yields the size of int. – pmg Sep 04 '19 at 16:37
Got that thanks for your response.. – Kannan Mrithunjai Sep 04 '19 at 17:38

What will be output of the below C code in 32-bit machine..?

1 Answers1