自己实现字符串转数字

来源：互联网发布：js输出对象的内容编辑：程序博客网时间：2024/06/05 17:56

A. atoi，字符串转整数

使用assert强制规定字符串一定是合法的，剩下的就是正负号了。

#include <stdio.h>#include <assert.h>int atoi(const char* str) {    if (str == NULL)        return 0;    int num = 0, i = 0;    bool neg = false;    if (str[0] == '-') {        neg = true;        ++i;    }    while (str[i] != '\0') {        assert (str[i] >= '0' && str[i] <= '9');        num *= 10;        num += int(str[i]-'0');        ++i;    }    if (neg)        return -num;    return num;}int main() {    printf("%d\n", atoi("233"));    printf("%d\n", atoi("980"));    printf("%d\n", atoi("066"));    printf("%d\n", atoi("-123"));    return 0;}

B. stof，字符串转float

这里只实现了简单的浮点数形式，没有实现像1e-10这样子的转化。
一开始输出的时候，以为是精度的问题，然后发现，转化的精度方面不会差太多，主要是float表示不了那么精确的浮点数，所以导致结果输出与原始数字有点不同：
233.0000000000, real=233.0000000000

3.1415927410, real=3.1415927410

123.4567871094, real=123.4567871094

-123.4567871094, real=-123.4567871094

#include <stdio.h>#include <assert.h>/*// version 1float stof(const char* str) {    float num = 0;    int i = 0;    for (i = 0; str[i] != '\0' && str[i] != '.'; ++i) {        assert(str[i] >= '0' && str[i] <= '9');        num *= 10;        num += int(str[i] - '0');    }    if (str[i] == '\0')        return num;    float digit = 0.0, div = 1.0;    for (++i; str[i] != '\0' && str[i] != '.'; ++i) {        digit *= 10.0f;        digit += int(str[i] - '0');        div *= 10.0f;    }    return digit/div + num;}*/// version 2: maybe more precisefloat stof(const char* str) {    if (str == NULL)        return 0.0f;    float num = 0;    int i = 0;    bool neg = false;    if (str[0] == '-') {        neg = true;        ++i;    }    while(str[i] != '\0' && str[i] != '.') {        assert(str[i] >= '0' && str[i] <= '9');        num *= 10;        num += int(str[i] - '0');        ++i;    }    if (str[i] == '\0')        return (neg) ? -num : num;    int start = i + 1, end = start;    while (str[end] != '\0' && str[end] != '.')        ++end;    float digit = 0.0f;    for (int j = end-1; j >= start; --j) {        assert(str[j] >= '0' && str[j] <= '9');        digit += int(str[j] - '0');        digit *= 0.1f;    }    return (neg) ? -(digit + num) : (digit + num);}int main() {    float a = 233.0f, b = 3.1415928f, c = 123.456789f;    printf("%.10f,\t real=%.10f\n\n", stof("233"), a);    printf("%.10f,\t real=%.10f\n\n", stof("3.1415928"), b);    printf("%.10f,\t real=%.10f\n\n", stof("123.456789"), c);    printf("%.10f,\t real=%.10f\n\n", stof("-123.456789"), -c);    return 0;}

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