课后练习题答案
Chapter 1
1.1 ×√×√√×√××√
1.2 b c
1.3 ad
1.4 semicolon printf math.h \n
Chapter 2
2.1 ×√××√√×√××√×
2.2 typedef 255 external const
Chapter 3
3.1 ×√××√√×××√√√
3.2 integer modula 6 logical sizeof paratheses typeconversion precedence
3.3 F F T F F F
3.4 F T T T F
3.5 (b) (c)
3.6 0 -2 7 10.25 false 3 3 1
3.10 0 1 1 1 1
3.11
d
}
100
3.12
110
111
3.13  1
3.14  200
3.15 x<=y
3.16 TRUE
3.19
2
1
4
3.20
-40
40
Chapter 4
4.1 ×√√√√√××√√√×
4.2 %hd %x ctype.h %l %*d [^] blank 6 - %e
4.4 (a) scanf(“%d %c %d”, &a, &b, &c);  (b)scanf(“%d %f %s”, &a, &b, &c);
(c) scanf(“%d-%d-%d”, &a, &b, &c);  (d) scanf(“%d %s %d”, &a, &b, &c);
4.5 (a)10x1.230000  (b)1234x 1.23      (c)1234    456.000000    (d) “123.40  ”    (e) 1020 1222(乱码)
4.7 （a）1988 x (b)乱码 （c）120 乱码 （d）乱码 x
4.8 （a）1275 -235.740000 
(b) 1275
-235.740000
(c) 0 0.000000
(d)      1275xxxx-235.74
(e)Cambridge
(f)1275      Cambridge
4.10  1988 无 无
Chapter 5
5.1 √×√××××××√
5.2 &&  switch break if-else x=y
5.4 （a）x = 2; y = 0; (b) x = 1; y = 0;
5.5 (a) if (grade <= 59)
            if (grade >= 50)
                second = second + 1;
(b) if (number > 100) printf(“out of range”);
else if (number < 0) printf(“out of range”);
else sum = sum + number;
(c)
if (T > 200)  printf(“admitted”);
else if (M > 60)
{if (M > 60) printf(“admitted”);}
else printf(“not admitted”);
5.6 F T F T
5.8 (a) x > 10 (b) (x != 10)||(y ! = 5) || (z >= 0)
(c) (x + y != z) || (z > 5)  (d) (x > 5) || (y != 10) || (z >= 5)
5.9 (a) x = 5; y = 10; z = 1    (b) x = 5; y = 10; z = 1 
(c) x = 5; y = 0; z =0  (d) 无变化
5.10  （a） x= 0; y = 2; z = 0;    (b) x = 1; y = 2; z = 0;
5.12    8
5.13  Delhi Bangalore END
5.14 
2
4
4
8
5.15 0 0 2
5.16  25
5.17 Number is negative
5.18 ABC
5.19 10
5.20 无输出
Chapter 6
6.1 √√√××√×√√×
6.2 n continue infinite indefinite-repetition-loop counter-variable
6.9 （a） 43210 (b)4321 (c)55555…55555(d)10 8
6.11 （a）无数次 x = 10; x = 5; x = 10; x = 5; x = 10;…
(b)5次 m = 3, m = 5, m = 7, m = 9, m = 11
(c)无数次, i 恒为0
(d)4次  m = 11 n = 9; m = 12 n = 11; m = 13, n = 13; m = 14, n = 15;
6.13
（a）for (n = 1; n <= 32; n = n * 2)  printf("%d ", n);
（b）for (n = 1; n <= 243; n = n * 3)  printf("%d ", n);
（c）for (n = -4; n <= 4; n = n + 2)  printf("%d ", n);
(d) sum = 0;
for (i = 2; i <= 16;  i  = sum)
    {
        sum = sum + i;
        n = n - i;
        printf("%d %d %d \n", i, sum, n);
    }
6.16 100 90 80 70 60 50 40 30 20 10 0
6.17 m = 20时陷入死循环， 程序无输出
6.18 1
c语言程序设计教材答案6.19 1
6.20 死循环，延长时间用
Chapter 7
7.1 √×√××√√√√××√
7.2 index/subscript run-time dynamic multi-dimensional sorting
7.14 15
7.15 HLOWRD
Chapter 8
8.1 ×√√√×√××√×××√×√
8.2 %[ a-z,A-Z] strcpy 3 stdlib gets strlen strstr strcmp s1-s2 puts
8.6 (a) The sky is the limit.
(b)                The sky is
(c) T
(d) The sky is the limit
(e)84
104
101
32
115
107
121
32
105
115
32
116
104
101
32
108
105
109
105
116
(f)输出1 2 3 4 5 6 7 8 9…21这些ASCII码对应的字符，无显示符号
(g)%
(h)U
8.7 (d)
8.8 7
8.9 (a)he(b)heorshe(c)5 7
8.11 pune
Chapter 9
9.1 √×××√×××√√√×××√×√√××
9.2 actual-parameter local-variable int data-type variable-name scope recursive static type auto
9.8 abceg
9.9 abcdeg
9.12 (a) 5 (b) 4 (c)3 (d)0
9.13 (a) 5 4 4 0 (b) 5.0 4.0 3.0 0.67
9.14 题目错，设z = 5；结果为：200 1000
9.15 100
9.16 x = 3; y = 0
Chapter 10
10.1 ×√√√√×√√×√√√××√
10.2 typedef  union  tag_name  pointer  member
Chapter 11
11.1 √×√√√√×√××
11.2 address * * 0 null

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