《C语言程序设计（Visual C++ 6.0环境）》电子教案：接本数据类型及运算符

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Email,qiaolin@cic.tsinghua.edu.cn
Tel,62792961
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– C
– C
– C
–
– C
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C
–
–
–
–
–
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–
–
–
–
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– 1~9 0~9 9 –876 55
– 0x” 0X” 0~9 A/a~F/f 0x55
0x1F 0x2a
– 0x5516 = 8510
– 0 0~7 055
– 0558 = 4510
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–
– 3.1416 –2.718281828
– E / e
– 3.14e3 = 3.14?103 –2.7E–2 = –2.7?10–2
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– A’ c’
– ASCII
– \
\n’
–
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0 1 2 3 4 5 6 7 8 9
0 \000 \001 \002 \003 \004 \005 \006 \a \b \t
10 \n \v \f \r \016 \017 \020 \021 \022 \023
20 \024 \025 \026 \027 \030 \031 \032 \033 \034 \035
30 \036 \037 space ! " # $ % & '
40 ( ) * +,–,/ 0 1
50 2 3 4 5 6 7 8 9,;
60 < = >? @ A B C D E
70 F G H I J K L M N O
80 P Q R S T U V W X Y
90 Z [ \ ] ^ _ ` a b c
100 d e f g h i j k l m
110 n o p q r s t u v w
120 x y z { | } ~ \177
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\a
\b
\f
\n
\r
\t
\v
\0 ASCII0
\\ \
\ \
\ \
\ddd ddd?ASCII
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T s i n g h u a \0
– Tsinghua” Programming in C”
– n n+1 NULL \0
– Tsinghua” 9
\084 \115 \105 \110 \103 \104 \117 \097
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– #define
– #define PI 3.14
–
–
–
–
#include <stdio.h>
#define PI 3.14
int main()
{
float r,area;
r = 3.0;
area = PI * r * r;
printf(?%f?,area);
return 0;
}
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– char int float double void
C
– CPU 4 8
– signed unsigned short long
– void
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char 8 –128 127
unsigned char 8 0 255
signed char/char 8 –128 127
int 32 –2,147,483,648 2,147,483,647
signed int/int 32 –2,147,483,648 2,147,483,647
unsigned int 32 0 4,294,967,295
short int/short 16 –32,768 32,767
signed short int/short 16 –32,768 32,767
unsigned short int 16 0 65,535
long int/long 32 –2,147,483,648 2,147,483,647
signed long int/long 32 –2,147,483,648 2,147,483,647
unsigned long int 32 0 4,294,967,295
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float 32 7 –3.40e38 –1.18e–38 1.18e–38 3.40e38
double 64 15 –1.80e308 –2.23e–308 2.23e–308 1.80e308
long double 80 19 –1.19e4932 –3.36e–4932 3.36e–4932 1.19e4932
void 32
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–
–
–
VANT
– value
– address
– name
– type
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#include <stdio.h>
int main()
{
char c;
int x,y,z;
unsigned long distance;
float m;
m = 1.11;
m = 2 * m;
m = m + 2.2;
printf(?%f\n?,m);
return 0;
}
x,y,z
c
distance
m
m
m
m
m
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char c = ‘c’;
float x,y,z = 4.53;
double a = 15.5,b = 2.0;
z4.53x,y
c c’ASCII99
a,b
–
–
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–
–
–
– 39 46
– 15
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15,[] () {} ++{} – –{}
14 sizeof ++{} – –{} ~ ! + – &{} *{} (){}
13 *{} / %
12 + –
11 << >>
10 < <= > >=
9 == !=
8 &{}
7 ^
6 |
5 &&
4 ||
3?,
2 = *= /= %= += –= <<= >>= &= ^= |=
1,
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–
–
–
floatdoublelong double
–
() b = c + (int)a;
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–
–
– while( ( c = getchar() ) != EOF ){ /* < */ }
–
– = b = c + (int)a;
–
–
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–
– y *= x + 1 y = y * (x + 1)
+= y += x; y = y + x;
–= y –= x; y = y – x;
*= y *= x; y = y * x;
/= y /= x; y = y / x;
%= y %= x; y = y % x;
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– + – * / %
–
–
–
– x = ( –b + sqrt( b * b – 4 * a * c ) ) / ( 2 * a )
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– > == < <= >= !=
–
–
– C 0 0 1
– x == y’a’ < ‘b’
– if( grade < 60 ) printf(?Fail\n?);
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– && || !
–
–
x && y x y
x || y x y
!x x
– x > y || ’a’ < ‘b’ && 3 <= 2
–
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– & | ^ ~ <<
>>
x y x & y x | y x ^ y ~y
0 0 0 0 0 1
0 1 0 1 1 0
1 0 0 1 1 1
1 1 1 1 0 0
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– x = x & y x 0xB675y 0x00FF
– x 16 88 0
x = 0xB675 1 0 11 0 11 0 10 1 01 1 10
y = 0x00FF 0 0 00 0 00 0 11 1 11 1 11
x = x & y 1 0 11 0 11 0 10 1 01 1 100 00 00
0
– 4 0 y = x & 0x1000 y 0 0 1
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1
– x = x | y x 0xB675y 0x00FF
– x 16818
x = 0xB675 1 0 11 0 11 0 10 1 01 1 10
y = 0x00FF 0 0 00 0 00 0 11 1 11 1 11
x = x | y 1 0 11 0 11 0 10 1 01 1 101 1 1
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– x = x ^ y x 0xB675y 0x00FF
– x 16818
x = 0xB675 1 0 11 0 11 0 10 1 01 1 10
y = 0x00FF 0 0 00 0 00 0 11 1 11 1 11
x = x ^ y 1 0 11 0 11 0 10 1 01 1 1001 0 10 0 01
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– x = ~x x 0xB675
x = 0xB675 1 0 11 0 11 0 10 1 01 1 10
x = ~x 1 0 11 0 11 0 10 1 01 1 100 1 00 1 00 1 01 0 10 0 01
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<< 2n0
– <<
– y = x << n x 0xB675n 2
x = 0xB675 1 0 11 0 11 0 10 1 01 1 10
y = x << n 1 0 11 0 11 0 10 1 01 1 101 0 1 00 1 1 0 0
>> 2n0
– >>
– y = x >> n x 0xB675n 2
x = 0xB675 1 0 11 0 11 0 10 1 01 1 10
y = x >> n 1 0 11 0 11 0 10 1 01 1 100 0 1 0 1 01 0 11
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– 1? 2,3
– 1 2
3
– x = ( a > b )? a,b;
– if( a > b ) x = a; else x = b;
–
–
– z = ( x = m + n,y = m – n,x / y );
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–
– C
– stdio.h”
–
–
–
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printf()
printf()
– printf(?,);
– printf(?age = %d\n?,age);
%y
– %d %i %f %e
%c %s %u %x
%o
– %[–]m[.n]y?m” %y”
.n” –
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#include <stdio.h>
int main()
{
int x,y,z;
x = 10;
y = 15;
z = 25;
printf(?Results as follows:\n?);
printf(?x = %d,y = %d,z = %d\n?,x,y,z);
printf(?x + y = %d\nx + y + z = %d\n?,x + y,x + y + z);
return 0;
}
Results as follows:
x = 10,y = 15,z = 25
x + y = 25
x + y + z = 50
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scanf()
scanf()
– scanf(?,);
– scanf(?%d,%c?,&age,&c);
–
–
–
Tab
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#include <stdio.h>
int main()
{
char c,s[10];
int i;
float x;
scanf(?%d,%c,%f,%s?,&i,&c,&x,&s);
printf(?i = %d,c = %c,x = %f,s = %s\n?,i,c,x,s);
return 0;
}
10,a,3.14,hello<>
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getchar()
–
–
– char c; c = getchar();
putchar(c)
–
– c
– char c = ‘c’; putchar(c);
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41
– 5 6 13 14
42
– 1 4