CodePudding user response:

 int main (int arg c, const char * argv []) {
Cout<" The first kind of (float) (2.2 1.2) to 1.0="& lt; <(float) (2.2 1.2) to 1.0 & lt; Cout<" The second: (float) (2.2-1.2-1.0)="& lt; <(float), (2.2-1.2-1.0) & lt; 
Int sf=sizeof (int) (float). 
Int sd=sizeof (int) (double); 
Printf (" sizeof (float)=% d \ n ", sf); 
Printf (" sizeof (double)=% d \ n ", sd); 

Double d1=float (2.2 1.2) to 1.0; 
Double d2=2.2-1.2-1.0; 
Float=f1 float (2.2 1.2) to 1.0; 
Float f2=float (2.2-1.2-1.0). 

Char * p=(char *) & amp; D1; 
int i; 
Printf (" d1="); 
for (i=0; iPrintf (" % 02 x, "(* (p + I) & amp; 0 x00ff)); 
} 
printf("\n"); 

P=(char *) & amp; D2. 
Printf (" d2="); 
for (i=0; iPrintf (" % 02 x, "(* (p + I) & amp; 0 x00ff)); 
} 
printf("\n"); 


P=(char *) & amp; F1; 
Printf (" f1="); 
for (i=0; iPrintf (" % 02 x, "(* (p + I) & amp; 0 x00ff)); 
} 
printf("\n");//f1 truncation d1 low four are all 0 

P=(char *) & amp; F2; 
Printf (" f2="); 
for (i=0; iPrintf (" % 02 x, "(* (p + I) & amp; 0 x00ff)); 
} 
printf("\n");//f2 truncation d2 low four is not zero, according to the floating point format (float format) output//memory format you want to view the floating-point 


return 0; 
}