1. Exercises at the upper limit of the classroom
3. Read the following program and write the running result:
analysis:
(1) The remainder of a is 0;
(2) b Take the quotient without rounding.
(3) c = b ++, execute c = b first, then execute b = b + 1.
(4) Double and float type data printing prints 6 digits after the decimal point by default. If you want to display more, you can use the form of% .nf.
(5) 15/10 belong to integer type, after taking the quotient, display 6 digits after the decimal point.
(6) d / 10 is an operation between float types, so it can display decimals.
answer:
#include <stdio.h>
int main(){
printf("0,2,1,15.000000,1.000000,1.500000");
return 0;
}
2. Exercises under the classroom
1. Assignment expressions and assignment statements, and write the results of the program
#include "stdio.h"
main()
{
printf("_______________________");
}
1) There are 256 ASCII characters from 0-255, c = 330, after 255, it will start from 0, 330-255 = 74, ASCII = 74, the character is J.
ASCII 为48，字符为0
ASCII 为65，字符为A
ASCII 为97，字符为a
2) \ 141 is the octal number character represented by 141. Converting this octal 141 to decimal is 97. Check the ASCII code, 97 is a, so the character a will be output.
'\101'表示ASCII为八进制101（对应十进制65）的字符，即 ‘ A ’ .
八进制转十进制 (101)8=(1*8^2+0*8^1+1*8^0）=（65）10
