c++ - 在 C++ 中从 'char' 到 'const Char*' 的无效转换

Question

我已经声明了一个函数(C++)

int products(char num1, char num2, char num3, char num4, char num5);

但是我的编译器给我这个错误:

Invalid conversion from 'char' to 'const Char*'
error initializing argument 1 of 'int atoi (const char*)'  on line 22

当我尝试将 num1、num2 等作为参数传递给 atoi 时。

#include<iostream>
#include<string>

using namespace std;

int Product(char num1, char num2, char num3, char num4, char num5);

const string LargeNum = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450";

int main(){
int greatestVal = 0;
for(int i = 0; i + 4 <= LargeNum.length(); i++ ){
    if(greatestVal < Product(LargeNum[i], LargeNum[i+1], LargeNum[i+2], LargeNum[i+3], LargeNum[i+4])){
        greatestVal = Product(LargeNum[i], LargeNum[i+1], LargeNum[i+2], LargeNum[i+3], LargeNum[i+4]);
    }
}
cout << greatestVal << endl;
system("PAUSE");
}

int Product(char num1, char num2, char num3, char num4, char num5){
    return (atoi(num1)*atoi(num2)*atoi(num3)*atoi(num4)*atoi(num5));
}

Answer 1

atoi 采用 const char*(即以 null 结尾的字符序列)。您为它提供了一个 char，所以编译器会报错。如何修复它完全取决于您要尝试做什么。

我猜您想将表示数字的字符转换为整数，“0”转换为 0，“1”转换为 1 等。如果是这样，那么正确的代码将是

return (num1 - '0')*(num2 - '0')*(num3 - '0')*(num4 - '0')*(num5 - '0');

之所以可行，是因为当您对它们进行算术运算时，字符会自动转换为整数，而且因为字符“0”到“9”保证按顺序排列，因此您只需将数字转换为整数从中减去“0”。