Discussion
What is the output of this C code?
#include <stdio.h>
void foo(float *);
int main()
{
int i = 10, *p = &i;
foo(&i);
}
void foo(float *p)
{
printf("%f\n", *p);
}
Answer: B
Correct answer: 0.000000
The address of an int is passed to a function expecting a float *, so the four bytes holding 10 are read as though they were an IEEE single-precision float.
The bit pattern 0x0000000A read that way is a subnormal number of about \( 1.4 \times 10^{-44} \), which %f prints as 0.000000. No conversion takes place - the same bits are simply reinterpreted. Note that the mismatched pointer types would draw a diagnostic in any modern compiler.
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