Solution for Exercise 19-3
ex1903
#include <stdio.h>
int main()
{
char c = 'c';
int i = 123;
float f = 98.6;
double d = 6.022E23;
printf("Address of 'c' %p\n",c);
printf("Address of 'i' %p\n",i);
printf("Address of 'f' %p\n",f);
printf("Address of 'd' %p\n",d);
return(0);
}
Output
Address of 'c' 0x63
Address of 'i' 0x7b
Address of 'f' 0x120a8
Address of 'd' 0x120a8
Notes
* The program compiles, but the output is suspect, of course.
* The warnings you see imply that a pointer value is expected when the %p conversion character is used.
* The type 'void *' in the warning message doesn't mean the void variable type was required. Instead, void * is how the compiler expresses its desire for a pointer.
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