C语言day11 字符串及其函数（二）

迷南。 2023-07-04 15:39 5阅读 0赞

## 用strcpy替换数组的一部分 ##

#include <stdio.h>
    #include <string.h>
    #define WORDS "beast"
    
    int main()
    { 
        char *orig = WORDS;
        char copy[40] = "Be the best that you can be.";
    
        puts(orig);
        puts(copy);
        strcpy(copy + 7, WORDS);
        puts(copy);
    
        return 0;
    }

beast
    Be the best that you can be.
    Be the beast

## strncpy（更安全） ##

strcpy 和strcat 都有一个安全漏洞  
![在这里插入图片描述][20200213142715405.png]  
之前用strcpy做的示例程序里，目标数组很大可以容纳源字符串，所以没出错，我把目标数组改小

#include <stdio.h>
    #include <string.h>
    #define SIZE 40
    #define LIM 5
    char *s_gets(char *st, int n);
    int main()
    { 
        char temp[SIZE];
        char target[LIM][LIM];
        int i=0;
        printf("Enter %d words beginning with q:\n", LIM);
        while(i<LIM && s_gets(temp, SIZE)!= NULL)//一定要把i<LIM放在前面，这样第五个q开头的单词输入后就直接退出了，用短路运算符时一定把首要判断条件放在前面，顺序是需要考虑一下的哦
        { 
            if(temp[0]!='q')// 等效于if(strncmp(temp, "q", 1))
            { 
                puts("This word is not beginning with q!");
            }
            else
            { 
                if(i!=4)
                    puts("Got one! Keep entering:");
                strcpy(target[i], temp);
                i++;
            }
    
        }
        printf("End of entering. The input words beginning with q:\n");
        for(i=0;i<LIM;i++)
            puts(target[i]);
        return 0;
    }char *s_gets(char *st, int n)
    { 
        char *ret_val;
        int i=0;
        ret_val = fgets(st, n, stdin);
        if(ret_val)
        { 
            while(*(st+i)!='\n' && st[i]!='\0')
                i++;
            if(st[i]=='\n')
                st[i] = '\0';
            else
                while(getchar()!='\n')
                    continue;
        }
        return ret_val;
    }

输出呈现这种错误，这是因为：  
输入question太长了，临时数组长度为40，存的下，但是复制给目标数组时，目标数组长度只有5，所以只复制了前5个字符，没有复制到空字符，第二个quarter又是如此····  
输出时，puts没有遇到空字符，就无法正确识别字符串结尾，于是输出了第一次遇到空字符前的所有字符，第二次输入也是如此，只是开始的位置变成了数组第二个元素的地址

Enter 5 words beginning with q:
    question
    Got one! Keep entering:
    quarter
    Got one! Keep entering:
    quality
    Got one! Keep entering:
    queen
    Got one! Keep entering:
    query
    End of entering. The input words beginning with q:
    questquartqualiqueenquery
    quartqualiqueenquery
    qualiqueenquery
    queenquery
    query

用strncpy

if(i!=4)
    	puts("Got one! Keep entering:");
    strncpy(target[i], temp, LIM-1);
    target[i][LIM-1]='\0';

Enter 5 words beginning with q:
    quarter
    Got one! Keep entering:
    quality
    Got one! Keep entering:
    question
    Got one! Keep entering:
    queen
    Got one! Keep entering:
    quote
    End of entering. The input words beginning with q:
    quar
    qual
    ques
    quee
    quot

# sprintf #

sprint 把数据写入字符串，多以可以把很多项（包括数字和字符串）合并为字符串  
printf写入屏幕  
sprintf第一个参数是目标字符串的地址，其余参数和printf相同，也是格式字符串和待写入项列表

#include <stdio.h>
    #include <string.h>
    #define MAX 10
    char *s_gets(char *st, int n);
    int main()
    { 
        char first[MAX], last[MAX], all[MAX*2+20];
        double prize;
    
        puts("Enter your first name:");
        s_gets(first, MAX);
        puts("Enter your last name:");
        s_gets(last, MAX);
        puts("Enter your prize:");
        scanf("%lf", &prize);//注意scanf输入double用%lf,输出double用%f
        sprintf(all, "%s %s $%.2f\n", last, first, prize);
        puts(all);
        return 0;
    }
    char *s_gets(char *st, int n)
    { 
        char *ret_val;
        int i=0;
        ret_val = fgets(st, n, stdin);
        if(ret_val)
        { 
            while(*(st+i)!='\n' && st[i]!='\0')
                i++;
            if(st[i]=='\n')
                st[i] = '\0';
            else
                while(getchar()!='\n')
                    continue;
        }
        return ret_val;
    }

Enter your first name:
    galler
    Enter your last name:
    mary
    Enter your prize:
    1000
    mary galler $1000.00

# 字符串函数总结 #

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70]![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 1]![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 2]![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 3]

## 示例 字符串排序 strcm 选择排序算法 交换指针！！！ ##

这个程序的巧妙之处是它通过交换指针实现排序

#include <stdio.h>
    #include <string.h>
    #define LIM 10
    #define SIZE 40
    char *s_gets(char *st, int n);
    //void st_sort(char st[][SIZE], int n);
    void st_sort(char *st[], int n);//指针数组的形式表示字符串数组
    int main()
    { 
        unsigned int ct=0;
        unsigned int i;
        char *ptr;
        char st[LIM][SIZE];//字符串数组就是二维数组
        char *ptstr[LIM];//[]优先级高于*,ptstr先和[]结合,所以ptstr是一个数组,
        //又和前面的*结合，成为指针数组，即元素是指针
    
        printf("Please enter up to %d lines, and I will sort them.\n", LIM);
        puts("Press the enter key to stop.");
        //最首要的判定条件放在最前面，即ct<LIM
        //s_gets会把换行符替换为空字符，所以如果不输满LIM个字符，按下换行符就退出
        while(ct<LIM && (ptr=s_gets(st[ct], SIZE))!=NULL && st[ct][0]!='\0')
        { 
            ptstr[ct] = st[ct];//把每个输入的字符串的地址存在指针数组中
            ct++;
        }
        st_sort(ptstr, ct);
        //输出排序后的字符串
        puts("The sorted strings:");
        for(i=0;i<ct;i++)
            puts(ptstr[i]);
        return 0;
    }
    char *s_gets(char *st, int n)
    { 
        char *ret_val;
        int i=0;
        ret_val = fgets(st, n, stdin);
        if(ret_val)
        { 
            while(*(st+i)!='\n' && st[i]!='\0')
                i++;
            if(st[i]=='\n')
                st[i] = '\0';
            else
                while(getchar()!='\n')
                    continue;
        }
        return ret_val;
    }
    
    //void st_sort(char st[][SIZE], int n)//数组法表示形参
    void st_sort(char *st[], int n)
    { 
        unsigned int top, seek;
        char *temp;
        for(top=0;top<n-1;top++)
        { 
            for(seek=top+1;seek<n;seek++)
            { 
                if(strcmp(st[top], st[seek])>0)//大于0则st[top]应排在st[seek]后面
                { 
                    temp = st[top];
                    //交换指针！！！
                    st[top] = st[seek];//如果用数组法表示形参，这句代码错误，数组和数组不可直接赋值，但是指针形参使得编译器认为他们都是指针，可以赋值
                    st[seek] = temp;
                }
            }
        }
    }

可以看到，strcmp的返回值的正负性特别有用，使得我们的字符串排序变得极其轻松

不过，写程序时，我发现自己的问题：

*  写关系表达式易错，本题是按下换行符则停止输入，我之前写成st\[ct\]\[0\]==’\\0’，结果程序不进去循环，我调试发现前两个表达式都没问题才发现自己写错了。易错
 *  排序算法中由于交换值会涉及到数组和数组之间的赋值，这在数组的使用规定中是不允许的，所以必须用指针法表示形参
 *  函数原型和函数定义的形参形式要一致，要么都是数组法，要么就都是指针法，不可混用，否则无法通过编译，报错

Please enter up to 10 lines, and I will sort them.
    Press the enter key to stop.
    fdfk
    uio
    op
    ih
    oa
    sdfg
    
    The sorted strings:
    fdfk
    ih
    oa
    op
    sdfg
    uio

### 选择排序 ###

这个排序算法我在草稿纸上推了，很简单，比如10个数字要从小到大排序：  
3 6 1 2 7 0 9 4 8 5  
首先top=0,st\[top\]=3，seek从1开始向右遍历，当seek=2时st\[seek\]更小，于是交换  
1 6 3 2 7 0 9 4 8 5  
top还是0，seek=3，直到seek=5再次达到交换条件  
0 6 3 2 7 1 9 4 8 5  
······  
所以第一次内层循环最多交换n-1次  
而第二次内层循环，top=1，则最多交换n-2次  
top=n-2时，最多交换一次  
（top取值范围：0—n-2，seek取值范围：top+1—n-1）

所以最多交换次数：  
 1 + 2 + 3 + ⋯ + n − 1 = n ( n − 1 ) 2 1+2+3+\\cdots+n-1=\\frac\{n(n-1)\}\{2\} 1\+2\+3\+⋯\+n−1=2n(n−1)  
![在这里插入图片描述][20200213182933511.png]

## 示例 计算标点符号个数（用到ctype.h中的函数） ##

toupper

ispunct是ctype.h中的函数  
![在这里插入图片描述][20200213192829314.png]

#include <stdio.h>
    #include <ctype.h>
    #include <string.h>
    #define LIMIT 101
    void ToUpper(char *st);
    unsigned int PunctCount(const char *);
    
    int main()
    { 
        char line[LIMIT];
        char *find;
    
        puts("enter a line.");
        fgets(line, LIMIT, stdin);
        find = strchr(line, '\n');//查找换行符，如果找到了，就返回换行符的地址，否则返回空指针
        if(find)//空指针的值是0
            *find = '\0';//用空字符替换换行符
        ToUpper(line);
        puts(line);
        printf("That line has %u punctuation characters.\n", PunctCount(line));
        return 0;
    }
    
    void ToUpper(char *st)
    { 
        while(*st)//不是空字符则进入循环
        { *st = toupper(*st);
        st++;}
    }
    
    unsigned int PunctCount(const char *st)
    { 
        unsigned int n=0;
        while(*st)
        { 
            if(ispunct(*st))
                n++;
            st++;
        }
        return n;
    }

这个程序值得学习的地方是while(\*st)和if(find)，利用了空字符的指针的值是0

enter a line.
    I'm trying to calculate the number of punctuation characters, and can you hele me?
    I'M TRYING TO CALCULATE THE NUMBER OF PUNCTUATION CHARACTERS, AND CAN YOU HELE ME?
    That line has 3 punctuation characters.

如果\#define LIMIT 81则

enter a line.
    I'm trying to calculate the number of punctuation characters, and can you hele me?
    I'M TRYING TO CALCULATE THE NUMBER OF PUNCTUATION CHARACTERS, AND CAN YOU HELE M
    That line has 2 punctuation characters.

# 给main函数带参数 #

## 示例1 输出参数和参数的个数 ##

这个示例是在命令行执行，输出参数的个数和参数

参数的个数是操作系统检测到后赋给main的参数argc的（c是count的缩写），OS还把所有参数都赋给了main的指针数组参数，即把所有参数的地址发给了main函数

#include <stdio.h>
    int main(int argc, char *argv[])
    { 
        int count;
    
        printf("The command line has %d arguments:\n", argc-1);
        for(count=1;count<argc;count++)
            printf("%d: %s\n", count, argv[count]);
        printf("\n");
    
        return 0;
    }

直接在编译器中执行，不行，因为无法输入参数

The command line has 0 arguments:
    
    
    Process returned 0 (0x0)   execution time : 0.597 s
    Press any key to continue.

在命令行执行，可以输入参数

C:\Users\wulimmya>cd C:\Users\wulimmya\Documents\MyCCode\MyC\bin\Debug
    
    C:\Users\wulimmya\Documents\MyCCode\MyC\bin\Debug>MyC Resistance is futile
    The command line has 3 arguments:
    1: Resistance
    2: is
    3: futile

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 4]![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 5]

## 示例2 把命令行参数字符串转换为数字 ##

这里说的是把键盘输入的数字，实际是一个个字符组成的字符串，把他们转换为数值形式，才可以用于程序的计算  
![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 6]

#include <stdio.h>
    #include <stdlib.h>
    int main(int argc, char *argv[])
    { 
        int i, times;
    
        if(argc<2 || (times=atoi(argv[1]))<1)
            printf("Usage: %s positive-number\n", argv[0]);
        else
            for(i=0;i<times;i++)
                puts("Hello, good looking!");
        return 0;
    }

atoi函数把字符串转换为int类型的整数

C:\Users\wulimmya>cd C:\Users\wulimmya\Documents\MyCCode\MyC\bin\Debug
    
    C:\Users\wulimmya\Documents\MyCCode\MyC\bin\Debug>MyC 3
    Hello, good looking!
    Hello, good looking!
    Hello, good looking!
    
    C:\Users\wulimmya\Documents\MyCCode\MyC\bin\Debug>MyC 1
    Hello, good looking!
    
    C:\Users\wulimmya\Documents\MyCCode\MyC\bin\Debug>MyC
    Usage: MyC positive-number

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 7]

#include <stdio.h>
    #include <stdlib.h>
    #define LIM 30
    char *s_gets(char *st, int n);
    int main()
    { 
        char number[LIM];
        char *end;
        long value;
    
        puts("Enter a number (empty line to quit):");
        while(s_gets(number, LIM) && number[0]!='\0')
        { 
            value = strtol(number, &end, 10);
            printf("base 10 input, base 10 output:%ld, stopped at %s (%d)\n",
                   value, end, *end);
    
            value = strtol(number, &end, 16);
            printf("base 16 input, base 10 output: %ld, stopped at %s (%d)\n",
                  value, end, *end);
            puts("Next number:");
    
        }
        puts("Bye!");
        return 0;
    }
    
    char *s_gets(char *st, int n)
    { 
        char *ret_val;
        int i=0;
        ret_val = fgets(st, n, stdin);
        if(ret_val)
        { 
            while(*(st+i)!='\n' && st[i]!='\0')
                i++;
            if(st[i]=='\n')
                st[i] = '\0';
            else
                while(getchar()!='\n')
                    continue;
        }
        return ret_val;
    }

Enter a number (empty line to quit):
    4
    base 10 input, base 10 output:4, stopped at  (0)
    base 16 input, base 10 output: 4, stopped at  (0)
    Next number:
    45
    base 10 input, base 10 output:45, stopped at  (0)
    base 16 input, base 10 output: 69, stopped at  (0)
    Next number:
    10atom
    base 10 input, base 10 output:10, stopped at atom (97)
    base 16 input, base 10 output: 266, stopped at tom (116)
    Next number:

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 8]![在这里插入图片描述][20200213203336136.png]![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzM2NjA3ODk0_size_16_color_FFFFFF_t_70 9]

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