The C programing language 第三章课后题-FinClip官网

The C programing language 第三章课后题

网友投稿 690 2022-09-16

The C programing language 第三章课后题

#include #include "chap1.c"/*这个是上一章的课后题,里面的函数,刚好这次需要,但是需要把chap1.c 中的main方面注释掉因为,c语言中,一个文件中只能有一个main方法*/#include "chap2.c" /* chap2.c 是第二章的课后题里面有的函数,这里需要使用使用 */#define abs(x) ((x)<0?-(x):(x)) /* 注意这里只是abs() 进行替换,如果看不懂可以看看这个/*3-3 : expand(s1,s2)*/int is_digit_letter(int c){ if(c >= 'a' && c <= 'z') return 1; else if(c >= '0' && c <= '9') return 1; return 0; }int expand_s2(char s2[], int start_point, int start, int end){ while(start <= end) s2[start_point++] = start++; s2[start_point] = '\0'; return start_point; //return s2 length }void expand(char s1[], char s2[]){ int i ,start , end; int index = 0; (s1[0] == '-')?(i=1,s1[0]='a'):(i=0); while(s1[i] != '\0') { if(is_digit_letter(s1[i])==1 && s1[i-1]=='-') { end = s1[i]; printf("end : %c : %d \n ", end, end); index = expand_s2(s2, index, start, end); printf("index : %c : %d \n ", index, index); if(s1[i+1] == '-') { start = ++end; printf("start : %c : %d \n ", start, start); } } else if(is_digit_letter(s1[i])) { start =s1[i]; printf("start : %c : %d \n ", start, start); } i++; }}/*看完答案之后,修改如下*/void expand_(char s1[], char s2[]){ char c; int i, j; i = j = 0; while((c=s1[i++]) != '\0') { if(is_digit_letter(c) && s1[i] == '-' && s1[i+1] >= c) { i++; while(c <= s1[i]) s2[j++] = c++; (s1[i+1] == '-')?(j--):' '; //处理 a-b-g ,重复出现b这种情况 } } s2[j] = '\0';}/*test_expand function: evaluating the expand function*/void test_expand(){ char s1[] = "a-z"; char s2[] = "a-c-g"; char s3[] = "a-c0-9"; char s4[] = "-a-c-g"; char s5[50]; expand_(s1,s5); printf("%s \n",s5); expand_(s2,s5); printf("%s \n",s5); expand_(s3,s5); printf("%s \n",s5); expand_(s4,s5); printf("%s \n",s5);}/* 3-4 首先,我们知道,有符号char 的取值范围: -128 到 127 同样的有符号的 int 取值范围是: 2^(sizeof(int)*8-1) 到 2^(sizeof(int)*8-1)-1 这个时候 : 2^(sizeof(int)*8-1) 这个值,在计算机输出的时候是一个负值,这个是最大负值其实在补码表示的时候,他的补码和反码一样,举个例子char -128 补码反码都是 1000 0000 所以,假如我们要把最大负数,转成字符串输出,就不能直接对n进行 n=-n 赋值操作,这样的操作计算的结果肯定有问题,我们把n=-n 赋值操作,间接的放到处理数据的过程中去,这样就不存在最大值的问题,因为处理数据的过程中,结果范围是 0-9 itoa : convert n to characters in s and can handle the largest negative numbner n = -2^(wordsize-1),regardless of the machine on which it runs*/void itoa(int n, char s[]) { int sign, i; i = 0; sign = n; do { s[i++] = abs(n%10) + '0'; }while(n /= 10); (sign<0)? (s[i++] = '-'):' '; s[i] = '\0'; reverse(s,_strlen(s)); }/*test_itoa evaluating the brove of function*/void test_itoa(){ char s[100]; int n = power(2,sizeof(int)*8-1); printf("%d \n",n); itoa(n,s); printf("%s \n",s); n = 0; printf("%d \n",n); itoa(n,s); printf("%s \n",s); n = 23; printf("%d \n",n); itoa(n,s); printf("%s \n",s);}/*3-5 itob function : converts the integer n into a base b character representation in the string s */void itob(int n, char s[], int b){ int i = 0; int sign = n; int temp; do{ temp = abs(n%b); s[i++]= (temp >9)? (temp+55): (temp+'0'); }while(n /= b); (sign < 0)? (s[i++]='-'): ' '; s[i] = '\0'; reverse(s,_strlen(s));}void test_itob(){ char s[55]; itob(18, s, 10); printf("%s \n",s); itoa(18, s); printf("%s \n",s); itob(18, s, 16); printf("%s \n",s); itob(-18, s, 8); printf("%s \n",s); itob(255, s, 16); printf("%s \n",s);}/*itoa : convert n to characters in sand can handle the largest negative numbnern = -2^(wordsize-1),regardless of the machine on which it runsminwidth is the least width of s that must be padded with blankson the left if necessary to make it wide enough*/void itoa_three(int n, char s[], int minwidth) { int sign, i; i = 0; sign = n; do { s[i++] = abs(n%10) + '0'; }while(n /= 10); (sign<0)? (s[i++] = '-'):' '; while(i < minwidth) s[i++] = ' '; s[i] = '\0'; reverse(s,_strlen(s)); }/* test_itoa evaluating the brove of function */void test_itoa_three(){ char s[100]; int n = power(2,sizeof(int)*8-1); printf("%d \n",n); itoa_three(n, s,1); printf("%s \n",s); itoa_three(n, s, 38); printf("%s \n",s);}main(){test_itoa_three();}

'; return start_point; //return s2 length }void expand(char s1[], char s2[]){ int i ,start , end; int index = 0; (s1[0] == '-')?(i=1,s1[0]='a'):(i=0); while(s1[i] != '

') { if(is_digit_letter(s1[i])==1 && s1[i-1]=='-') { end = s1[i]; printf("end : %c : %d \n ", end, end); index = expand_s2(s2, index, start, end); printf("index : %c : %d \n ", index, index); if(s1[i+1] == '-') { start = ++end; printf("start : %c : %d \n ", start, start); } } else if(is_digit_letter(s1[i])) { start =s1[i]; printf("start : %c : %d \n ", start, start); } i++; }}/*看完答案之后,修改如下*/void expand_(char s1[], char s2[]){ char c; int i, j; i = j = 0; while((c=s1[i++]) != '

') { if(is_digit_letter(c) && s1[i] == '-' && s1[i+1] >= c) { i++; while(c <= s1[i]) s2[j++] = c++; (s1[i+1] == '-')?(j--):' '; //处理 a-b-g ,重复出现b这种情况 } } s2[j] = '

';}/*test_expand function: evaluating the expand function*/void test_expand(){ char s1[] = "a-z"; char s2[] = "a-c-g"; char s3[] = "a-c0-9"; char s4[] = "-a-c-g"; char s5[50]; expand_(s1,s5); printf("%s \n",s5); expand_(s2,s5); printf("%s \n",s5); expand_(s3,s5); printf("%s \n",s5); expand_(s4,s5); printf("%s \n",s5);}/* 3-4 首先,我们知道,有符号char 的取值范围: -128 到 127 同样的有符号的 int 取值范围是: 2^(sizeof(int)*8-1) 到 2^(sizeof(int)*8-1)-1 这个时候 : 2^(sizeof(int)*8-1) 这个值,在计算机输出的时候是一个负值,这个是最大负值其实在补码表示的时候,他的补码和反码一样,举个例子char -128 补码反码都是 1000 0000 所以,假如我们要把最大负数,转成字符串输出,就不能直接对n进行 n=-n 赋值操作,这样的操作计算的结果肯定有问题,我们把n=-n 赋值操作,间接的放到处理数据的过程中去,这样就不存在最大值的问题,因为处理数据的过程中,结果范围是 0-9 itoa : convert n to characters in s and can handle the largest negative numbner n = -2^(wordsize-1),regardless of the machine on which it runs*/void itoa(int n, char s[]) { int sign, i; i = 0; sign = n; do { s[i++] = abs(n%10) + '0'; }while(n /= 10); (sign<0)? (s[i++] = '-'):' '; s[i] = '

'; reverse(s,_strlen(s)); }/*test_itoa evaluating the brove of function*/void test_itoa(){ char s[100]; int n = power(2,sizeof(int)*8-1); printf("%d \n",n); itoa(n,s); printf("%s \n",s); n = 0; printf("%d \n",n); itoa(n,s); printf("%s \n",s); n = 23; printf("%d \n",n); itoa(n,s); printf("%s \n",s);}/*3-5 itob function : converts the integer n into a base b character representation in the string s */void itob(int n, char s[], int b){ int i = 0; int sign = n; int temp; do{ temp = abs(n%b); s[i++]= (temp >9)? (temp+55): (temp+'0'); }while(n /= b); (sign < 0)? (s[i++]='-'): ' '; s[i] = '

'; reverse(s,_strlen(s));}void test_itob(){ char s[55]; itob(18, s, 10); printf("%s \n",s); itoa(18, s); printf("%s \n",s); itob(18, s, 16); printf("%s \n",s); itob(-18, s, 8); printf("%s \n",s); itob(255, s, 16); printf("%s \n",s);}/*itoa : convert n to characters in sand can handle the largest negative numbnern = -2^(wordsize-1),regardless of the machine on which it runsminwidth is the least width of s that must be padded with blankson the left if necessary to make it wide enough*/void itoa_three(int n, char s[], int minwidth) { int sign, i; i = 0; sign = n; do { s[i++] = abs(n%10) + '0'; }while(n /= 10); (sign<0)? (s[i++] = '-'):' '; while(i < minwidth) s[i++] = ' '; s[i] = '

'; reverse(s,_strlen(s)); }/* test_itoa evaluating the brove of function */void test_itoa_three(){ char s[100]; int n = power(2,sizeof(int)*8-1); printf("%d \n",n); itoa_three(n, s,1); printf("%s \n",s); itoa_three(n, s, 38); printf("%s \n",s);}main(){test_itoa_three();}

随着系统集成技术的发展，如何利用小程序容器技术提升企业的数字化转型效率是现代企业面临的重要挑战。企业应关注如何实现跨平台兼容、数据安全管理以及全生命周期管理，确保运营合规与灵活应变，以适应不断变化的市场需求和技术环境。

690 2022-09-16

The C programing language 第三章课后题

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