[C Supplement] Conversion of Integer to String

一、库函数

1.1 atoi字符串转换成int型整数

1.1.1 函数原型

int atoi(const char* str);

• The parameter is the string address to be converted;
• The function returns the converted integer.

1.1.2 函数说明

• 函数返回 int 类型,So for exceeding 10 位的字符串,The conversion is bound to fail, 10Bits can also go wrong,因为 int 的最大值为 2,147,483,647 .
• The function conversion process is：从左到右,Skip whitespace characters first,Find the digit character after the whitespace character,Other non-numeric characters after numeric characters are ignored.If the first non-whitespace character in the string is not a valid integer,或str为空,or contain only whitespace characters,函数返回0

1.1.3 使用举例

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

int main(){
    

    char str1[11] = {
    0}, str2[11] = {
    0}, str3[11] = {
    0}, str4[15] = {
    0};
    int num1, num2, num3, num4;
    strcpy(str1, " 12snd34f.6");
    strcpy(str2, " ab12,st");
    strcpy(str3, "1111111111");
    strcpy(str4, "11111111111");
    num1 = atoi(str1);
    num2 = atoi(str2);
    num3 = atoi(str3);
    printf("%d\n%d\n%d\n%d\n", num1, num2, num3, num4);  

    return 0;
}

输出结果：

(The last bit of data overflowed)

1.1.4 strtol字符串转换成长整型(long)

1.2 itoa整数转换成字符串

1.2.1 函数原型

char* itoa(int value, char* str, int redix);

• itoa 全称 integer to ASCII
• 参数表：
  • value：The integer value to convert
  • str：The storage address of the converted string
  • redix：基数,Represents the base number of an integer

1.2.2 函数说明

• itoa函数头文件为stdlib.h;
• 如果取10进制,且 value 为负数,A minus sign is added before the conversion result(-),The rest of the bases are treated as unsigned;
• Cardinality means will value Convert to the corresponding base number and then convert to a string.
• Since the function parameter is int类型,Therefore the converted value cannot exceedint的存储范围.
• The character array to be stored also needs to be long enough to hold the converted numeric string;
• The function returns the address of the string.

注意： 与atoi不同,This function is a non-standard library function,为Windows独有,Some compilers may not support this function,To write cross-platform programs,用更为强大的sprintf函数即可,用法举例：sprintf(str, “%x”, 100); 将100以16The base form is converted to a string and stored in str指向的内存区域.

use control：

itoa	sprintf
itoa(value, str, 10);	sprintf(str, "%d", value);
itoa(value, str, 8);	sprintf(str, "%o", value);
itoa(value, str, 16);	sprintf(str, "%x", value);

1.2.3 使用举例

#include<stdio.h>
#include<stdlib.h>


int main(){
    

    char str[1000] = {
    0};
    const int mlen = 11;			//最高位数
    int num = 1;

    for(int i=1; i<=mlen; i++){
    
        itoa(num, str, 10);			//以10进制转换为字符串
        printf("%10s", str);
        itoa(num, str, 16);			//以16进制转换为字符串
        printf(" %s\n", str);
        num += pow(10, i);
    }
    
    return 0;
}

输出结果：

• On the left is a decimal string,右侧为16进制字符串;
• The last data overflowed,超出 int 存储范围.

二、大数转换

• For the above library functions only int The data in the range is valid,If out of range,other methods are required.

2.1 字符串转换成整数

• For pure numeric strings only,使用最大的long long 类型

#include<stdio.h>
#include<string.h>

long long my_atoi(char* str);

int main(){
    

    char str[20] = {
    0};
    scanf("%s", str);

    long long n = my_atoi(str);
    printf("%lld\n", n);

    return 0;
}

long long my_atoi(char* str)
{
    
    int i = strlen(str) - 1;
    long long result = 0, base = 1, a = 0;
    for( ; i>=0; i--){
    
        a = (str[i] - '0')*base;            //Multiply each digit by the corresponding number of digits,个位乘1,十位乘10...
        result += a;
        base *= 10;
    }

    return result;
}

2.2 整数转换成字符串

char* my_itoa(int n)
{
    
    int tmp_n = n, len = 0;     //数字的位数
    while(tmp_n){
    
        tmp_n /= 10;
        len++;
    }

    char* result = (char*)malloc(sizeof(char)*(len+1));
    if(n == 0) strcpy(result, "0");
    else{
    
        for(int i=len-1; i>=0; i--){
    
            result[i] = n%10 + '0';
            n /= 10;
        }
        result[len] = '\0';
    }
    
    return result;
}

三、 综合应用 —— 大数相加

题目要求：
Read two numbers as strings,Add and output as a string.

思路：
A bit-by-bit addition method is used,从最低位开始,Convert each character to a number and add them,记录进位的值,Convert the sum of each bit to character storage,Finally, reverse the sequence of characters.

char* solve(char* s, char* t ) {
    

    int i, j;
    i = strlen(s) - 1;			//从最低位开始
    j = strlen(t) - 1;
    int max = i>j ? i : j ;		//Allocate the character array length,考虑进位,The longest is the highest digit plus1
    char* result = (char*)malloc(sizeof(char)*(max + 3));
    int len = 0;                 //The number of digits in the summed value
    int sum = 0, add = 0;        //各位之和,进位值
    //从低到高,逐位相加
    while(i>=0 || j>= 0 || add != 0){
    
        int x = i >= 0 ? s[i] - '0' : 0;
        int y = j >= 0 ? t[j] - '0' : 0;
        
        sum = x + y + add;
        result[len++] = sum % 10 + '0';
        add = sum/10;
        
        i--;
        j--;
    }
    //Numeric characters are reversed
    for(i=0; 2*i < len; i++){
    
        char tmp = result[i];
        result[i] = result[len - i - 1];
        result[len - i - 1] = tmp;
    }
    result[len] = 0;
    
    return result;
}