atof(将字符串转换成浮点型数)
atoi(将字符串转换成整型数)
atol(将字符串转换成长整型数)
strtod(将字符串转换成浮点数)
strtol(将字符串转换成长整型数)
strtoul(将字符串转换成无符号长整型数)
toascii(将整型数转换成合法的ASCII 码字符)
toupper(将小写字母转换成大写字母)
tolower(将大写字母转换成小写字母)
1 int atoi(const char *nptr) 2 { 3 int c; /* current char */ 4 int total; /* current total */ 5 int sign; /* if '-', then negative, otherwise positive */ 6 7 /* skip whitespace */ 8 while ( isspace((int)(unsigned char)*nptr) ) 9 ++nptr; 10 11 c = (int)(unsigned char)*nptr++; 12 sign = c; /* save sign indication */ 13 if (c == '-' || c == '+') 14 c = (int)(unsigned char)*nptr++; /* skip sign */ 15 16 total = 0; 17 18 while (isdigit(c)) { 19 total = 10 * total + (c - '0'); /* accumulate digit */ 20 c = (int)(unsigned char)*nptr++; /* get next char */ 21 } 22 23 if (sign == '-') 24 return -total; 25 else 26 return total; /* return result, negated if necessary */ 27 }
1 double atof(char s[]) //将字符串s转换成double型的浮点数 2 { 3 4 int i; 5 int sign; 6 int flag; 7 int expn; 8 double val; 9 double power; 10 11 sign = 1; 12 flag = 0; 13 power = 1.0; 14 expn = 0; 15 for(i = 0; isspace(s[i]); ++i) 16 ; 17 if(s[i] == '-') 18 sign = -1; 19 if(s[i] == '+' || s[i] == '-') 20 ++i; 21 for(val = 0.0; isdigit(s[i]); ++i) 22 val = val * 10.0 + (s[i] - '0'); 23 if(s[i] == '.') 24 ++i; 25 for(; isdigit(s[i]); ++i) 26 { 27 val = val * 10.0 + (s[i] - '0'); 28 //power = power * 10.0; 29 ++flag; 30 } 31 if(s[i] == 'e' || s[i] == 'E') //如果写成s[i++] == 'e' || s[i++] == 'E'，if(s[i] == '-') 32 //则当输入的字符串带有E时，不能正确得到结果，这是因为在一 33 //个语句中使用两次自增操作，引起歧义 34 if(s[++i] == '-') 35 { 36 ++i; 37 for(; isdigit(s[i]); ++i) 38 expn = expn * 10 + (s[i] - '0'); 39 expn = expn + flag; 40 power = pow(10, expn); 41 return sign * val / power; 42 } 43 else 44 { 45 for(; isdigit(s[i]); ++i) 46 expn = expn * 10 + (s[i] - '0'); 47 expn = expn - flag; 48 power = pow(10, expn); 49 return sign * val * power; 50 } 51 52 power = pow(10, flag); 53 return sign * val / power; 54 55 }