test of program.exe stopped working

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test of program.exe stopped working

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#1 by (3 votes)
#2 by (2 votes)

2

I'm new in the world of programming, I'm just starting, and it turns out that today when I run a simple program I jump "test program.exe has stopped working." I do not understand why I skipped this, it had never happened to me before and it seems that the problem is precisely with this part of the code. Could someone explain to me what happens? Thank you very much.

#include<stdio.h>
int main(){
    int dia;
    int auxd;
    int contd;

contd=0;
auxd=1;
    do{
        printf("ingrese dia: ");    scanf("%i",dia);

    if(auxd==dia){
        contd++;
        if (contd>4){
            printf(" el dia %i ha pasado las 4 consultas",dia);
        }
    }
    else{
        auxd=dia;
        contd=0;
    }
}while  (dia != 0);
 return 0;

}

c++

asked by Rich Sanchez 18.08.2018 в 02:12

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2 answers

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score 3 · Answer 1

As indicated in the comments, scanf needs a pointer to store the data entered by the user. However, if you label the question as C + + the logical thing is that you happen to use C ++ code and stop programming in C.

It would also be advisable that you tabulate the code correctly. Your program of less than 30 lines already begins to be difficult to read because it is wrongly tabulated ... when you have to do one of 300 lines it's going to hurt, really.

#include<iostream>

int main(){
  int contd = 0;
  int auxd = 0;
  int dia;

  do{
    std::cout << "ingrese dia: ";
    std::cin >> dia;

    if(auxd==dia){
      contd++;

      if (contd>4){
        std::cout << " el dia " << dia << " ha pasado las 4 consultas";
      }
    }
    else{
      auxd=dia;
      contd=0;
    }
  }while (dia != 0);
  return 0;
}

score 2 · Answer 2

It has already been indicated in comments and in the eferion response that the trigger for your failure is to misuse the scanf function. I join the advice of the rest of users when recommending that you read the manual of said function; However, knowing where the program fails does not explain why it fails. The explanation is somewhat complex.

The function scanf is part of the libraries of c and therefore follows the rules of the C language in its operation; One of the limitations of this language is the absence of references 1 and this limitation means that to modify a variable outside its scope it is necessary to pass the memory address of it.

Since you have passed "%i" as a read format, the scanf function will wait to write to an integer ( int ) and therefore assumes that it will receive an integer pointer. I insist on the concept that scanf assumes that it will receive an integer pointer , this function (from the C libraries) has no way to verify that the type of data provided matches the data type expected 2 and therefore assumes that it will be; It is the programmer's job to ensure that this precondition 3 .

In your code, you failed in your responsibility as a programmer to ensure the expected precondition for scanf , so he assumed that the integer you were passing as a parameter was an integer pointer. In your code we can see that you do not initialize variable dia , according to C ++ initialization rules this will leave said variable with an indeterminate value and that value will be what scanf interprets as a memory address that when trying to write in it a segment violation will occur and the operating system will inform you that your program.exe has stopped work .

You can know the difference between pointer and reference in this thread .

In other words: you can not guarantee security of the types .

For these reasons the scanf function is considered as unsafe, prone to errors and a possible security hole in the system, even if there is tutorials to take advantage of the vulnerabilities caused by the scanf function. Therefore in C ++ its use is discouraged and in C it is advised to use secure alternatives such as sscanf (< em> s ecure scanf ).