字符数组倒序

? 1 2 3 4 5 6 7 8 9 10 11 #include <stdio.h> void daoxu(char str[]){ int i; char temp; for(i = 0; i < strlen(str) / 2 ; i ++){ temp = str[i]; str[i] = str[strlen(str) - i-1]; str[strlen(str) - i-1] = temp; } }

单词计数

? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 int wordCount(char str[]){ int i; int count = 0; int word = 0; for (i = 0 ; str[i] != '\0' ; i ++) { if (str[i] == ' ') { word = 0; }else if (word == 0) { word = 1; count ++; } } return count; }

字符大写转小写

? 1 2 3 4 5 6 7 8 9 10 void mylwr(char str[]){ int i; for (i = 0 ; str[i] != '\0' ; i ++) { if (str[i] >= 'A' && str[i] <= 'Z') { str[i] += 'a' - 'A'; } } }

字符小写转大写

? 1 2 3 4 5 6 7 8 9 10 void myupr(char str[]){ int i; for (i = 0 ; str[i] != '\0' ; i ++) { if (str[i] >= 'a' && str[i] <= 'z') { str[i] -= 'a' - 'A'; } } }

字符数组计算字符串长度

? 1 2 3 4 5 int mylen(char str[]){ int len; for (len = 0 ; str[len] != '\0' ; len ++); return len; }

字符串连接

? 1 2 3 4 5 6 7 8 9 void mycat(char str1[],char str2[]){ int i,j; for (i = 0 ; str1[i] != '\0' ;i++); for (j = 0 ; str2[j] != '\0' ; j ++) { str1[i + j] = str2[j]; } str1[i + j] = '\0'; }

指定长度串接

? 1 2 3 4 5 6 7 8 9 void myncat(char str1[],char str2[], int len){ int i,j; for(i = 0; str1[i] != '\0'; i++); for (j = 0; j < len; j++) { str1[i + j] = str2[j]; } str1[i + j] = '\0'; }

字符数组拷贝

? 1 2 3 4 5 6 7 void mycpy(char dst[],char src[]){ int i = 0; do { dst[i] = src[i]; } while (src[i++] != '\0'); }

字符数组指定长度拷贝

? 1 2 3 4 5 6 7 8 void myncpy(char dst[],char src[], int len){ int i; for (i = 0; i < len; i++) { dst[i] = src[i]; } dst[i] = '\0'; }

找出句子中最长的单词

? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 void longest(char dst[],char src[]){ int i = 0,j; int count =0; int max = -1; do { if (src[i] ==' ' || src[i] == '\0') { if (count > max) { max = count; for (j = 0; j < count; j++) { dst[j] = src[i - count + j]; } dst[j] = '\0'; } count = 0; }else{ count++; } } while (src[i++] != '\0'); }

从字符串中提取整形数字

? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 #include <stdio.h> int getint(char str[], int a[]){//从字符串中提取数字并放在数组中 int i = 0; int w = 0; int c = 0; int j, k; do { if (str[i] > '0' && str[i] <= '9') { w++; }else if (w) { j = 0; for (k = w; k > 0; k--) { j *= 10; j += str[i - k] - '0'; } w = 0; a[c] = j; c++; } } while (str[i++] != '\0'); return c; } void main(){ char str[100]; int a[100]; int i, j; gets(str); i = getint(str,a); for (j = 0; j < i; j++) { printf("%d ",a[j]); } }

整形、字符数组型转换

? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 #include <stdio.h> #include <stdlib.h> int sumof1(int x)//求一个数转换成二进制以后1的个数 { int countx = 0; while(x) { countx ++; x &= x-1; //每位与一次x - 1；就能消掉最后一个1 } return countx; } void main(){ char c[10]; int i = 999; itoa(i, c, 10);//以10进制转换成字符数组 puts(c); itoa(i, c, 16);//以16进制转换成字符数组 printf("0x%s\n", c); itoa(i, c, 8);//以8进制转换成字符数组 printf("0%s\n", c); itoa(i, c, 2);//以2进制转换成字符数组 puts(c); i = atoi(c);//再将字符串转成整形 printf("%d\n", i); printf("%d\n", sumof1(i));//以2进制表示时1的个数 }