学完C语言之后,我就去阅读《C Primer Plus》这本经典的C语言书籍,对每一章的编程练习题都做了相关的解答,仅仅代表着我个人的解答思路,如有错误,请各位大佬帮忙点出!
由于使用的是命令行参数常用于linux系统或者vscode,但此代码是运行于vs2022的,测试截图就不弄了。
1.编写一个函数,把二进制字符串转换为一个数值。例如,有下面的语 句:
char * pbin = "01001001";
那么把pbin作为参数传递给该函数后,它应该返回一个int类型的值25。
#include <stdio.h>
#include <string.h>
int convert(const char* str)
{
int sum = 0, len = strlen(str);
for (int i = len - 1, exp = 1; i >= 0; --i, exp *= 2)
{
sum += (str[i] - '0') * exp;
}
return sum;
}
int main(void)
{
const char* pbin = "01001001";
printf("binary %s is decimal %d\n", pbin, convert(pbin));
return 0;
}
2.编写一个程序,通过命令行参数读取两个二进制字符串,对这两个二 进制数使用~运算符、&运算符、|运算符和^运算符,并以二进制字符串形 式打印结果(如果无法使用命令行环境,可以通过交互式让程序读取字符 串)。
#include <stdio.h>
#include <stdlib.h>
int bstr_to_dec(const char* str)
{
int val = 0;
while (*str)
{
val = val * 2 + (*str++) - '0';
}
return val;
}
char* itobs(int n, char* str)
{
int sz = 8 * sizeof(int);
for (int i = sz - 1; i >= 0; --i, n >>= 1)
{
str[i] = (n & 1) + '0';
}
str[sz] = '\0';
return str;
}
int main(int argc, char* argv[])
{
char bstr[8 * sizeof(int) + 1];
char str1[20] = {0};
char str2[20] = { 0 };
printf("请输入两个字符串:");
scanf("%s%s", str1, str2);
int v1 = bstr_to_dec(str1);
int v2 = bstr_to_dec(str2);
printf("~%s = %s\n", str1, itobs(~v1, bstr));
printf("~%s = %s\n", str2, itobs(~v2, bstr));
printf("%s & %s = %s\n", str1, str2, itobs(v1 & v2, bstr));
printf("%s | %s = %s\n", str1, str2, itobs(v1 | v2, bstr));
printf("%s ^ %s = %s\n", str1, str2, itobs(v1 ^ v2, bstr));
return 0;
}
3.编写一个函数,接受一个 int 类型的参数,并返回该参数中打开位的 数量。在一个程序中测试该函数。
#include <stdio.h>
char* itobs(int n, char* str)
{
int sz = 8 * sizeof(int);
for (int i = sz - 1; i >= 0; --i, n >>= 1)
{
str[i] = (n & 1) + '0';
}
str[sz] = '\0';
return str;
}
int onbits(int x)
{
int sum = 0;
for (int i = 8 * sizeof(int) - 1; i >= 0; --i)
{
sum += (x >> i) & 1;
}
return sum;
}
int main(int argc, char* argv[])
{
int x = 11;
char bstr[8 * sizeof(int) + 1];
printf("%d(%s) has %d bit on.\n", x, itobs(x, bstr), onbits(x));
return 0;
}
4.编写一个程序,接受两个int类型的参数:一个是值;一个是位的位 置。如果指定位的位置为1,该函数返回1;否则返回0。在一个程序中测试 该函数。
#include <stdio.h>
int test_bit(int x, int pos)
{
return (x >> pos) & 1;
}
char* itobs(int n, char* str)
{
int sz = 8 * sizeof(int);
for (int i = sz - 1; i >= 0; --i, n >>= 1)
{
str[i] = (n & 1) + '0';
}
str[sz] = '\0';
return str;
}
int main(int argc, char* argv[])
{
int x = 11;
int sz = 8 * sizeof(int);
char bstr[8 * sizeof(int) + 1];
printf("%d(%s):\n", x, itobs(x, bstr));
for (int i = sz - 1; i >= 0; --i)
{
printf("The position %d is %d\n", i + 1, test_bit(x, i));
}
return 0;
}
5.编写一个函数,把一个 unsigned int 类型值中的所有位向左旋转指定数 量的位。例如,rotate_l(x, 4)把x中所有位向左移动4个位置,而且从最左端 移出的位会重新出现在右端。也就是说,把高阶位移出的位放入低阶位。在 一个程序中测试该函数。
#include <stdio.h>
char* itobs(int n, char* str)
{
int sz = 8 * sizeof(int);
for (int i = sz - 1; i >= 0; --i, n >>= 1)
{
str[i] = (n & 1) + '0';
}
str[sz] = '\0';
return str;
}
unsigned int rotate_l(unsigned int n, unsigned int b)
{
static const int size = 8 * sizeof(int);
b %= size;
return (n << b) | (n >> (size - b));
}
int main(int argc, char* argv[])
{
unsigned int val = 11;
unsigned int rot = rotate_l(val, 4);
char bstr1[8 * sizeof(int) + 1];
char bstr2[8 * sizeof(int) + 1];
printf("%u rotated is %u.\n", val, rot);
printf("%s rotated is %s.\n", itobs(val, bstr1), itobs(rot, bstr2));
return 0;
}
6.设计一个位字段结构以储存下面的信息。
字体ID:0~255之间的一个数;
字体大小:0~127之间的一个数;
对齐:0~2之间的一个数,表示左对齐、居中、右对齐;
加粗:开(1)或闭(0);
斜体:开(1)或闭(0);
在一个程序中使用该结构来打印字体参数,并使用循环菜单来让用户改 变参数。例如,该程序的一个运行示例如下:
该程序要使用按位与运算符(&)和合适的掩码来把字体ID和字体大小 信息转换到指定的范围内。
#include <stdio.h>
#include <ctype.h>
#include <string.h>
typedef unsigned int uint;
typedef struct
{
uint id : 8;
uint sz : 7;
uint at : 2;
uint b : 1;
uint i : 1;
uint u : 1;
} font;
static font ft = { 1, 12, 0, 0, 0, 0 };
const char* state[4] = { "off", "on" };
const char* alignment[7] = { "left", "center", "right" };
void eatline(void)
{
while (getchar() != '\n')
continue;
}
int get_first(void)
{
int ch;
do
{
ch = getchar();
} while (isspace(ch));
eatline();
return ch;
}
int get_choice(void)
{
int ch;
printf("ID SIZE ALIGNMENT B I U\n");
printf("%-7u%-9u%-12s", ft.id, ft.sz, alignment[ft.at]);
printf("%-8s%-8s%-8s\n", state[ft.b], state[ft.i], state[ft.u]);
printf("f) change font s) change size a) change alignment\n");
printf("b) toggle bold i) toggle italic u) toggle underline\n");
printf("q) quit\n");
while (ch = get_first(), NULL == strchr("fsabiuq", ch))
{
printf("Please enter with f, s, a, b, i, u or q: ");
}
return ch;
}
void change_font(void)
{
int ch;
uint id;
printf("Enter font id (0-255): ");
while (scanf("%u", &id) != 1)
{
while ((ch = getchar()) != '\n')
{
putchar(ch);
}
printf(" is not a id.\n");
printf("Please enter a number such as 0, 5 or 255: ");
}
ft.id = id & 0XFF;
}
void change_size(void)
{
int ch;
uint sz;
printf("Enter font sz (0-127): ");
while (scanf("%u", &sz) != 1)
{
while ((ch = getchar()) != '\n')
{
putchar(ch);
}
printf(" is not a size.\n");
printf("Please enter a number such as 0, 5 or 127: ");
}
ft.sz = sz & 0x7F;
}
void change_alignment(void)
{
int ch;
printf("Select alignment:\n");
printf("l) left c) center r) right\n");
while (ch = get_first(), NULL == strchr("lcr", ch))
{
printf("Please enter with l, c or r: ");
}
ft.at = (ch == 'l' ? 0 : ch == 'c' ? 1 : 2);
}
void change_toggle(int ch)
{
if (ch == 'b')
{
ft.b ^= 1;
}
else if (ch == 'i')
{
ft.i ^= 1;
}
else
{
ft.u ^= 1;
}
}
int main(void)
{
int ch;
while ((ch = get_choice()) != 'q')
{
switch (ch)
{
case 'f':
{
change_font();
break;
}
case 's':
{
change_size();
break;
}
case 'a':
{
change_alignment();
break;
}
case 'b':
case 'i':
case 'u':
{
change_toggle(ch);
break;
}
}
putchar('\n');
}
printf("Bye!\n");
return 0;
}
7.编写一个与编程练习 6 功能相同的程序,使用 unsigned long 类型的变 量储存字体信息,并且使用按位运算符而不是位成员来管理这些信息。
#include <stdio.h>
#include <ctype.h>
#include <string.h>
typedef unsigned long ulong;
static ulong ft = 0x00001180;
const char* state[4] = { "off", "on" };
const char* alignment[7] = { "left", "center", "right" };
void eatline(void)
{
while (getchar() != '\n')
continue;
}
int get_first(void)
{
int ch;
do
{
ch = getchar();
} while (isspace(ch));
eatline();
return ch;
}
int get_choice(void)
{
int ch;
printf("ID SIZE ALIGNMENT B I U\n");
printf("%-7u%-9u%-12s", (ft >> 12) & 0XFF, (ft >> 5) & 0x7F, alignment[(ft >> 3) & 0x03]);
printf("%-8s%-8s%-8s\n", state[(ft >> 2) & 1], state[(ft >> 1) & 1], state[ft & 1]);
printf("f) change font s) change size a) change alignment\n");
printf("b) toggle bold i) toggle italic u) toggle underline\n");
printf("q) quit\n");
while (ch = get_first(), NULL == strchr("fsabiuq", ch))
{
printf("Please enter with f, s, a, b, i, u or q: ");
}
return ch;
}
void change_font(void)
{
int ch;
ulong id;
printf("Enter font id (0-255): ");
while (scanf("%lu", &id) != 1)
{
while ((ch = getchar()) != '\n')
{
putchar(ch);
}
printf(" is not a id.\n");
printf("Please enter a number such as 0, 5 or 255: ");
}
id &= 0XFF, id <<= 12;
for (int i = 12; i < 20; ++i)
{
ft &= ~(ulong)(1 << i);
}
ft |= id;
}
void change_size(void)
{
int ch;
ulong sz;
printf("Enter font sz (0-127): ");
while (scanf("%lu", &sz) != 1)
{
while ((ch = getchar()) != '\n')
{
putchar(ch);
}
printf(" is not a size.\n");
printf("Please enter a number such as 0, 5 or 127: ");
}
sz &= 0X7F, sz <<= 5;
for (int i = 5; i < 12; ++i)
{
ft &= ~(ulong)(1 << i);
}
ft |= sz;
}
void change_alignment(void)
{
int ch;
printf("Select alignment:\n");
printf("l) left c) center r) right\n");
while (ch = get_first(), NULL == strchr("lcr", ch))
{
printf("Please enter with l, c or r: ");
}
ft &= ~(ulong)(1 << 3), ft &= ~(ulong)(1 << 4);
ft = ft | (ch == 'c' ? (ulong)(1 << 3) : ch == 'r' ? (ulong)(1 << 4) : 0);
}
void change_toggle(int ch)
{
if (ch == 'b')
{
if (ft & 0x04)
{
ft &= ~(ulong)(0x04);
}
else
{
ft |= (ulong)(0x04);
}
}
else if (ch == 'i')
{
if (ft & 0x02)
{
ft &= ~(ulong)(0x02);
}
else
{
ft |= (ulong)(0x02);
}
}
else
{
if (ft & 0x01)
{
ft &= ~(ulong)(0x01);
}
else
{
ft |= (ulong)(0x01);
}
}
}
int main(void)
{
int ch;
while ((ch = get_choice()) != 'q')
{
switch (ch)
{
case 'f':
{
change_font();
break;
}
case 's':
{
change_size();
break;
}
case 'a':
{
change_alignment();
break;
}
case 'b':
case 'i':
case 'u':
{
change_toggle(ch);
break;
}
}
putchar('\n');
}
printf("Bye!\n");
return 0;
}