尽管是项目Euler计划,但以下代码实际上并没有太多关注。我想添加50个100位数字,我将每个数字的每个数字分配给数组加数[100] [50]中的元素。然后我会单独添加每个数字/位置,并携带额外的数字。这些数字是从名为Input.txt
的文本文件中读取的,它只包含所有数字。 http://projecteuler.net/problem=13
我无法从文件输入流(string numbers[100][50]
)向字符串数组(<fstream>
)的元素分配字符。问题在评论中有更完整的描述:
“[对于第一个循环]这个循环为字符串数组中的每个字符串分配一个数字。即使第二个数字(50)没有做任何事情(它似乎被std :: string覆盖;参见变量声明),它需要在那里循环才能工作。循环的“逻辑”相同;“j”什么都不做,但需要在那里让循环工作?“
而且,(对于第二个循环)“此循环从相应的字符串数组元素填充”addends [100] [50]“数组。如果我尝试使用数组”数字“调用”char_to_int()“ [i] [j]“,编译器抱怨输入不是正确的数据类型。添加变量”k“使循环工作一次,但最终在第二次循环崩溃(使用”numbers [i ] [j] [k]“)。所以我尝试了”char_to_int((numbers [i] [j])。c_str())“,但编译器抱怨”const char *“与”char“不兼容。一个指针解决了这个问题(“char_to_int(*((numbers [i] [j])。c_str()))”),但程序仍然会崩溃。“我拿出了一些无关紧要的代码,以使其更具可读性。
#include <iostream>
#include <fstream>
#include <string>
using namespace std;
int char_to_int(char chInput);
int main()
{
int placeholder; //so console doesn't close immediately upon finish
int sum[102] = {0}; // 100+2, 100 places + 2 places from carrying over
int addends[100][50] = {0};
string numbers[100][50];
ifstream input("Input.txt");
/* This loop assigns a number to every string in the string array. Even
* though the second number (50) doesn't do anything (it seems to be
* overridden by std::string; see variable declaration), it needs to be
* there for the loop to work. Same "logic" for the loop; "j" doesn't do
* anything but needs to be there??? Confused :-\
*/
for (int i = 0; i < 100; i++)
for (int j = 0; j < 1; j++)
getline(input, numbers[i][j]);
/* This loop fills in the "addends[100][50]" array from the corresponding
* string array element. If I try to call "char_to_int()" with the array
* "numbers[i][j]", the compliler complains that the input isn't of the
* right data type. Adding a variable "k" makes the loop work for one run,
* but eventually crashes on the second loop (using "numbers[i][j][k]").
* So I tried "char_to_int((numbers[i][j]).c_str())", but the compiler
* complains that "const char *" is incompatible with "char". Adding a
* pointer resolves the issue ("char_to_int( *( (numbers[i][j]).c_str() ) )"),
* but the program still crashes on the second loop through.
*/
for (int i = 0; i < 100; i++)
for (int j = 0; j < 50; j++)
for (int k = 0; k < 1; k++) //used when the variable "k" was being used
addends[i][j] = char_to_int( (numbers[i][j]).c_str() );
return 0;
}
代码未完成;我决定反对继续,因为我(显然)需要先解决这个问题。
答案 0 :(得分:3)
使用
编译并运行良好string numbers[100];
for (int i = 0; i < 100; i++)
getline(input, numbers[i]);
for (int i = 0; i < 100; i++)
for (int j = 0; j < 50; j++)
addends[i][j] = char_to_int( (numbers[i][j]));
删除stdafx.h
包含并定义char_to_int
。
std::string
本身包含一个字符数组,因此您只需要一个std::string
s的一维数组。然后,您可以通过[]
索引
numbers[i][j]
获取数组numbers
中第i个字符串的第j个字符(字节,而不是字节)。