Question

因此，我正在创建一个程序，该程序读入名为"Bankfile.txt"的文件，第一个数字“3”表示我们正在使用多少帐户。数字“123”“467”和“499”是银行帐号，每个号码旁边的数字是其原始余额。在我的代码中，我使用2D数组扫描它们。我认为我在扫描它们时有一切正确但是当我运行程序时，帐号被打印得非常奇怪，如here所示。关于为什么要这样打印的任何想法？

谢谢！

#include <stdio.h>


int main(){
FILE* file;
file = fopen("bankfile.txt","r");
int accounts,b=1,w=2,d=3,u=4,i,j,accNum,origBal;
float interest = .0185; 
float test;
float accountInfo[accNum][origBal];

fscanf(file, "%d", &accounts);

for(i = 0; i < accounts; i++)
      {
      fscanf(file, "%d", &accountInfo[i]);
      for(j = 0; j < 1; j++)
            {
            fscanf(file, "%f", &accountInfo[i][j]);
             printf("Account %d has a balance of $%.2f\n", accountInfo[i], accountInfo[i][j]);
            }

      } 

system("pause");
return 0;
}

Answer 1

好的，这里有不 二维数组 - 这在概念上是错误的。帐号只有一个余额。所以你所拥有的只是一个维度，但你的数据有几个字段...这就是你在C中使用function eval_force() % eval_force.m IS USED FOR EVALUATING FORCE % THE STRATEGY USUALLY ADOPTED FOR A LENNARD-JONES OR AS A MATTER OF FACT % ANY PAIR-WISE INTERACTING SYSTEM IS AS FOLLOWS: % 1. EVALUATE THE DISTANCE BETWEEN TWO PAIRS OF ATOMS % 2. ENSURE THAT MINIMUM IMAGE CONVENTION (MIC) IS FOLLOWED % 3. IF THE DISTANCE OBTAINED THROUGH MIC IS GREATER THAN THE CUT OFF % DISTANCE MOVE TO NEXT PAIR % 4. ELSE EVALUATE POTENTIAL ENERGY AND CALCULATE FORCE COMPONENTS % 5. F(i,j) = -F(j,i) global MASS KB TEMPERATURE NUM_ATOMS LENGTH TSTEP; global EPS SIG R_CUT GAMMA POT_E; global POSITION VELOCITY FORCE STO; dr = zeros(3,1); drh = zeros(3,1); FORCE(:) = 0.0; POT_E = 0.0; for ( i=1:NUM_ATOMS ) for ( j=i+1:NUM_ATOMS ) dist2 = 0.0; % VARIABLE dist2 STORES DISTANCE BETWEEN PAIR (i,j) % FIRST FIND OUT THE DIFFERENCE IN X,Y AND Z COORDINATES % VARIABLE dr IS USED FOR THIS PURPOSE for(k = 1:3) dr(k) = POSITION(i,k) - POSITION(j,k); % THESE STEPS ENSURE MINIMUM IMAGE CONVENTION IS FOLLOWED if(dr(k) > LENGTH/2.0) dr(k) = dr(k) - LENGTH; end if(dr(k) < -LENGTH/2.0) dr(k) = dr(k) + LENGTH; end % MINIMUM IMAGE CONVENTION ENDS HERE dist2 = dist2 + dr(k)*dr(k); % dist2 IS BASED UPON MIC end if(dist2 <= R_CUT*R_CUT) % IF THE CUT OFF CRITERIA IS SATISFIED dist2i = power(SIG,2)/dist2; dist6i = power(dist2i,3); dist12i = power(dist6i,2); POT_E = POT_E + EPS * (dist12i - 2*dist6i) + 33.34 * EPS * power(sqrt(dist2) - SIG,2)/(2 * power(SIG,2)); % STORES THE POTENTIAL ENERGY Ff = 12.0 * EPS * (dist12i-dist6i) - 33.34 * EPS * (sqrt(dist2) - SIG)/(dist2i * sqrt(dist2) * power(SIG,2)); Ff = Ff * dist2i; for(k = 1:3) FORCE(i,k) = FORCE(i,k) + Ff*dr(k)- GAMMA*VELOCITY(i,k); FORCE(j,k) = FORCE(j,k) - Ff*dr(k)- GAMMA*VELOCITY(j,k); end end end end end的地方。这里有一些示例代码可以产生你期望的输出：

struct

首先阅读第一行，然后根据需要只分配#include <stdio.h> #include <stdlib.h> #include <string.h> /* data structure holding a bank account */ typedef struct account { int id; double balance; } account; int main(void) { int cap = 1024; /* initial capacity of account array */ int count = 0; /* number of accounts */ char buf[1024]; /* buffer for a line of text */ char *tok; /* token from text line */ FILE *bankfile; int i; account *accounts = malloc(cap * sizeof(account)); bankfile = fopen("bankfile.txt", "r"); while (fgets(buf, 1024, bankfile)) { int accId; tok = strtok(buf, " \t\r\n"); if (!tok) continue; accId = atoi(tok); if (accId > 0) { /* first token in line was a positive integer */ tok = strtok(0, " \t\r\n"); if (tok) { /* there was a second token in the same line, then we found * an account with a balance */ accounts[count].id = accId; accounts[count].balance = atof(tok); if (++count == cap) { cap *= 2; accounts = realloc(accounts, cap * sizeof(account)); } } } } fclose(bankfile); for (i = 0; i < count; ++i) { printf("Account %d has a balance of $%.2f\n", accounts[i].id, accounts[i].balance); } free(accounts); return 0; }的元素，可以简化这一过程。

_{当然，对于制作，将错误检查添加到account和fopen()以及朋友......}

Answer 2

首先，我认为你不了解二维数组的概念。当你有一个2D数组（foo）并添加两个数组时，它将看起来像这样。

int foo[][] = { {1,2,3} {4,5,6} };

当你致电foo[0]时，它实际上会引用您添加的第一个数组（即[1,2,3]），而foo[1]将引用第二个数组。您应该使用两个名为accountNum和accountBal的单独数组。

最后，您永远不会给accNum或origBal赋予值，这会使您的2D数组成为空数组。因此，您可以根据accounts变量为它们动态分配内存。

记住这一点，你应该改变你的主要逻辑来解决这些问题。

fscanf(file, "%d", &accounts);
accountNum = malloc(sizeof(int) * accounts);
accountBal = malloc(sizeof(float) * accounts);

for(i = 0; i < accounts; i++) {
  fscanf(file, "%d", &accountNum[i]);
  fscanf(file, "%f", &accountBal[i]);
  printf("Account %d has a balance of $%.2f\n", accountNum[i], accountBal[i]);
}

free(accountBal);
free(accountNum);

二维阵列打印

2 个答案: