学习总结
1、if…else…从语义上看就能出用途，跟其他语言没差多少，只需要记住，世界上最遥远的距离之一：我走if你却走else。
2、根据个人几年的编程经验，太多的if…else…嵌套会加大代码的可读性和维护难度。个人认为代码最好不要超过三层if…else…的嵌套，否则最好使用布尔值控制流程。
3、逻辑运算符优先级：!>&&>||
4、运行到continue语句将导致剩余的迭代部分被忽略，开始下一次迭代。continue仅用于循环，而break语句用于循环和switch中。
5、编程题（题1）：
1 #include <stdio.h>
2
3 int main(){
4 int space=0,newline=0,other=0;
5 char ch;
6 printf("please enter something:\n");
7 while((ch=getchar())!=‘#‘){
8 if(ch==‘\n‘){
9 newline+=1;
10 }else if(ch==‘ ‘){
11 space+=1;
12 }else{
13 other+=1;
14 }
15 }
16 printf("space is %d\n",space);
17 printf("newline is %d\n",newline);
18 printf("other is %d\n",other);
19 return 0;
20 }
运行结果：
please enter something:
hello world!
hi nihao.
#ABC
space is 2
newline is 2
other is 19
6、编程题（题11）：
1 #include <stdio.h>
2 #define ARTICHOKE_UNIT_PRIC 1.25
3 #define BEET_UNIT_PRICE 0.65
4 #define CAROTA_UNIT_PRICE 0.89
5 #define DISCOUNT 0.05
6 #define T_0_5 3.50
7 #define T_5_20 10.00
8 #define T_20_ 0.1
9
10 int main(){
11 double a,b,c,ap,bp,cp,ac,bc,cc,sc,dc,tc;
12 ap=ARTICHOKE_UNIT_PRIC;
13 bp=BEET_UNIT_PRICE;
14 cp=CAROTA_UNIT_PRICE;
15 printf("how many artichoke you want(pound):");
16 scanf("%lf",&a);
17 if(a==0)return 0;
18
19 printf("how many beet you want(pound):");
20 scanf("%lf",&b);
21 if(b==0)return 0;
22
23 printf("how many carota you want(pound):");
24 scanf("%lf",&c);
25 if(c==0)return 0;
26
27 printf("\n------UNIT PRICE------\n");
28 printf("artichoke‘s unit price is $%.2f(one pound) \n",ap);
29 printf("beet‘s unit price is $%.2f(one pound)\n",bp);
30 printf("carota‘unit price is $%.2f(one pound)\n",cp);
31
32 printf("\n------ORDER------\n");
33 printf("artichoke:%.2fpound\n",a);
34 printf("beet:%.2fpound\n",b);
35 printf("carota:%.2fpound\n",c);
36
37 printf("\nartichoke is $%.2f",a*ap);
38 printf("\nbeet is $%.2f",b*bp);
39 printf("\ncarota is $%.2f\n",c*cp);
40 sc=a*ap+b*bp+c*cp;
41 printf("\ntotal cost is $%.2f",sc);
42 dc=sc>100?sc*DISCOUNT:0;
43 printf("\ndiscount is $%.2f",dc);
44
45 printf("\ntotal weight is %.2f",a+b+c);
46 if(0<(a+b+c)<=5){
47 tc=T_0_5;
48 }
49 if(5<(a+b+c) && (a+b+c)<=20){
50 tc=T_5_20;
51 }
52 if((a+b+c)>20){
53 tc=8+(a+b+c)*0.1;
54 }
55 printf("\nttransport cost is $%.2f",tc);
56 printf("\norder cost is $%.2f\n",sc-dc+tc);
57
58 return 0;
59 }
运行结果：
how many artichoke you want(pound):123
how many beet you want(pound):234
how many carota you want(pound):343
------UNIT PRICE------
artichoke‘s unit price is $1.25(one pound)
beet‘s unit price is $0.65(one pound)
carota‘unit price is $0.89(one pound)
------ORDER------
artichoke:123.00pound
beet:234.00pound
carota:343.00pound
artichoke is $153.75
beet is $152.10
carota is $305.27
total cost is $611.12
discount is $30.56
total weight is 700.00
ttransport cost is $78.00
order cost is $658.56
【C语言学习】《C Primer Plus》第7章 C控制语句：分支与跳转
原文：http://www.cnblogs.com/zhangyingai/p/7087387.html