编译错误发生在 C 中

Question

data02.in is numerical data in which one real number value is described in one line. From this numerical data, create a program that discovers the maximum and minimum values. The program to create is

1. It works even if the number of numbers stored in the file is changed.

2. Each time numerical data is read from a file, it is compared and updated with the maximum and minimum values.

It shall be. Create this according to the following procedure.

1. First, prepare three variables of double type that store the numerical value, maximum value, and minimum value to be read. The initial values ​​of the maximum value and the minimum value are set to the minimum value and the maximum value that can be taken by the numerical data, respectively. Since the given numerical data is known to be non-negative, the initial value of the maximum value is set to 0. The initial value of the minimum value is the maximum value of double type DBL_MAX. DBL_MAX is defined in the header file <float.h>, so you need to include it before you can use it. (DBL_MAX is a macro that represents the maximum value that can be represented by double type).
2. Read a number from a file. Compare this with the maximum value, and if it is larger, update the maximum value. The minimum value is also compared and updated in the same way.
3. Return to 2. and repeat the process until the next numerical value cannot be read.
4. Outputs the found maximum and minimum numbers up to the fourth decimal place to the file data02.out, and terminates the program.

data02.in

0.04352715661330464
0.5711499219614311
0.5510152700437206
0.4349183348106874
0.5655036481971432
0.778936411144494
0.9653850342847657
0.3212329719743001
0.14875984255792618
0.7334184832076945
0.3015893694560843
0.778596472495796
0.6062732756985176
0.8703907121064157
0.726552074397092
0.33992379312230137
0.1631129989837965
0.9676628599119076
0.585461667958953
0.8615281807925766
0.8007898389888695
0.9492303287750206
0.45691105334861526
0.11234397028174314
0.12211793272128402
0.5159930438804902
0.025879785403056355
0.19976268362579752
0.48253690657173065
0.7636664896336439

Error compile

    gcc prog02.c
prog02.c: In function ‘main’:
prog02.c:8:16: error: ‘MAX’ undeclared (first use in this function)
    8 |     double min=MAX;
      |                ^~~
prog02.c:8:16: note: each undeclared identifier is reported only once for each function it appears in
prog02.c:22:44: error: called object ‘max’ is not a function or function pointer
   22 |   fprintf(fpout,"max = %.3f, min = %.3f\n",max(n),min(n));
      |                                            ^~~
prog02.c:7:12: note: declared here
    7 |     double max=0;
      |            ^~~
prog02.c:22:51: error: called object ‘min’ is not a function or function pointer
   22 |   fprintf(fpout,"max = %.3f, min = %.3f\n",max(n),min(n));
      |                                                   ^~~
prog02.c:8:12: note: declared here
    8 |     double min=MAX;

source

#include <stdio.h>
#include <float.h>

int main(){
  FILE *fpin,*fpout;
  double n;
    double max=0;
    double min=MAX;
  
  fpin = fopen("rand.in","r");
  
  fpout = fopen("rand.out","w");

  while(fscanf(fpin,"%lf",&n) != EOF){
    
    if(n>max) max=n;
    
    if(n<min) min=n;
    
  }

  fprintf(fpout,"max = %.3f, min = %.3f\n",max(n),min(n));
    
  fclose(fpout);
  fclose(fpin);
}

Answer 1

There are several problems:

• MAX isn't declared anywhere. You could as well use FOOBAR , the result would be the same. BTW you were told to use DBL_MAX .
• max(n),min(n) : you're using max and min as functions just use max and min .

So you probably want something like this:

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

int main(){
  FILE *fpin,*fpout;
  double n;
  double max=0;
  double min= DBL_MAX;
  
  fpin = fopen("rand.in","r");
  if (fpin == NULL)
  {
    printf("Can't open file\n");
    exit(1);
  }
  
  fpout = fopen("rand.out","w");
  if (fpout == NULL)
  {
    printf("Can't create file\n");
    exit(1);
  }

  while(fscanf(fpin,"%lf", &n) != EOF){    
    if (n > max) max = n;    
    if (n < min) min = n;
  }

  fprintf(fpout,"max = %.3f, min = %.3f\n", max, min);
    
  fclose(fpout);
  fclose(fpin);
}