我已经在我的代码（无法正常工作）上浇注了一段时间。这是针对Project Euler问题的，其中找到一个非常大的和，然后要求打印出该和的前十个数字。 （可以在这里找到问题：https://projecteuler.net/problem=13）
我运行了几个“测试”，在这些测试中添加了打印命令，以查看代码中各个点的各种值。运行代码时，我得到了从符号到十位数的任何数字，应该是一位数字。
无论如何。我的问题是：这是类型转换问题还是我缺少的方法存在其他明显问题？我一直在研究类型转换，试图找到解决方法，但无济于事。
感谢您的任何帮助！
代码如下：
// this is a program to find a very large sum of many very large numbers
#include <stdio.h>
#include <math.h>
int main()
{
//declare all ints needed
int i;
int j;
int d; // digit, need to add 48
int placesum; // sum of addition in _'s place (1's, 10's, 10000's)
int place; // final place value
int c = 0, tens = 1, otherc; // counters for start finder
int a = 0; // another counter
//declare all arrays
char numarray[101][51]; //array of strings containing all 100 numbers
char sum[100];
printf("please save data to largesumdata.txt\n\n press enter when ready");
getchar();
// THE PROBLEM- I don't know how to get my data into my program // FIXED
// using fscanf()
FILE *pf; // declare a pointer to the file
pf = fopen("largesumdata.txt", "r"); // trys to open file // "r" means read only
if(pf == NULL)
printf("Unable to open file, sorry Jar\n");
else
{
for(j = 0; j < 100; j++)
fscanf(pf, "%s\n", &numarray[j]); // fscanf(pointer, data type, location)
}
//TESTING
//printf("You have reached point A\n");//POINT A WAS REACHED
//TESTING
//TESTING
//printf("Check1, %c\n", numarray[45][23]);
//TESTING
//TESTING
//printf("%c\n", numarray[90][22]);//Can successfully call characters from array
//TESTING
// (Brute force attempt) //I NEVER MESS WITH numarray WHY IS IT CHANGING
for(i = 49; i >= 0; i--)
{
//printf("%d\n", d);
for(j = 0; j < 100; j++)
{
d = (int)numarray[j][i] - 'o';
//printf("%d\n", d);
//holdup// d -= 48; // ASCII conversion // could also write "d = d-48"
//printf("%d\n", d);
placesum += d; // could also write "placesum = placesum + d"
//printf("%d\n", placesum);
}
place = placesum % 10;
placesum = placesum / 10; // takes "10's place" digit for next column
// now need to put 'int place' into 'char sum'
sum[i+5] = (char)place+'0'; // ASCII conversion // "+5" for extra space //HERE not properly stored in sum
}
//TESTING
//printf("Check2, %c\n", numarray[45][23]);
//TESTING
//TESTING
//printf("You have reached point B\n");//POINT B WAS REACHED
//TESTING
// find out where sum starts
for(c=0; c<10; c++)
if(sum[c] != '0')
break;
//TESTING
//printf("You have reached point C\n"); //POINT C WAS REACHED
//TESTING
otherc = 4-c;
printf("The first 10 digits of the sum of all those f***ing numbers is....\n");
printf("%d-%d-%d-%d-%d-%d-%d-%d-%d-%d", sum[otherc, otherc+1, otherc+2, otherc+3, otherc+4, otherc+5, otherc+6, otherc+7, otherc+8, otherc+9]);
//%c-%c-%c-%c-%c-%c-%c-%c-%c-%c //copy and paste purposes
//%d-%d-%d-%d-%d-%d-%d-%d-%d-%d // ^^^^^
getchar();
return 0;
}
附言如果我过多的笔记令人困惑，我深表歉意
最佳答案
您使用错误的格式在C中打印数组。sum[otherc, otherc+1, otherc+2, otherc+3, otherc+4, otherc+5, otherc+6, otherc+7, otherc+8, otherc+9]->这实际上衰减为sum[otherc+9]，因为C将,视为运算符。
要在每个数组索引处打印值，您应该像这样使用它：sum[otherc], sum[otherc+1], sum[otherc+2],..
要了解有关C的,（逗号）运算符的更多信息，可以begin here
如上所述，在您的printf中，第一个格式说明符%d为sum[otherc + 9]，因为sum[otherc,...,otherc+9]实际上是一个数字，并且是数组otherc + 9的第sum个索引。您没有为其他格式说明符提供要打印的任何内容，因此会产生垃圾。