I get segmentation error

This is a discussion on I get segmentation error within the C Programming forums, part of the General Programming Boards category; This program asks user to enter numbers between 1 and 100. After i run it i get segmentation error. The ...

  1. #1
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    Question I get segmentation error

    This program asks user to enter numbers between 1 and 100. After i run it i get segmentation error. The program doesnt repeat number twice. Can someone help me.
    Code:
    #include<stdio.h>
    #define SIZE 20
    int main(void)
    {
     int a[SIZE];
     int i;
     int x;
     int j;
     int flag;
     int idx=0;
    
    
      printf("Enter 20 numbers between 1 and 100\n");
      scanf("%d",&a[i]);
    
    //this is how to accept multiple numbers
      for (int i = 0; i < 20;i++){
            scanf("%d",&x);
            a[i]=x;
    
            // this is to determine that numbers dont repeat
            if(i==0){
                    a[i]=x;
                    idx++;
            }
            else{
                    for( int j=0; j<idx; j++){
    
    
                    }
    
            }
    
            if(flag !=1){
                    a[idx]=x;
                    idx++;
            }
    
      printf("The no duplicate values are:\n");
      printf("\n%d",x);
     }
    
    
    
    return 0;
    }

  2. #2
    Epy
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    Code:
    scanf("%d",&a[i]);
    You need to initialize i before you can use it. You initialize it after this point in your for loop.

  3. #3
    Epy
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    The error is probably coming from this though:

    Code:
    for (int i = 0; i < 20;i++){
    you already declared i as an int before the for loop, this should just be i = 0 (actually it shouldn't since you've already taken one value for your a array )

    You also declare j twice.

  4. #4
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    can you specify, what you mean?

  5. #5
    Epy
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    Code:
     int a[SIZE];
     int i;
     int x;
     int j;
     int flag;
     int idx=0;
    
    
      printf("Enter 20 numbers between 1 and 100\n");
      scanf("%d",&a[i]);
    You have not given a value to i, so where is scanf going to place the inputted value? &a[i] needs a value for i.

    Code:
    for (int i = 0; i < 20;i++){
    ...
    for( int j=0; j<idx; j++){
    You have already declared these variables at the beginning:

    Code:
     int a[SIZE];
     int i;
     int x;
     int j;
     int flag;
     int idx=0;
    you don't need to declare them again, just give them values, i.e. i = 0, j = 0

  6. #6
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    Like this... i still keep getting the segmentation error
    Code:
    #include<stdio.h>
    #define SIZE 20
    
    int main(void)
    {
     int a[SIZE];
     int i;
     int x;
     int j;
     int flag;
     int idx=0;
    
    
      printf("Enter 20 numbers between 10 and 100\n");
      scanf("%d",&a[i]);
    
    //this is how to accept multiple numbers
      for (i = 0; i < 20;i++){
            scanf("%d",&x);
            a[i]=x;
    
            // this is to determine that numbers dont repeat
            if(i==0){
                    a[i]=x;
                    idx++;
            }
            else{
                    for( j=0; j<idx; j++){
    
    
                    }
    
            }
    
            if (a[j]==x){
                    flag=1;
            }
    
            if(flag !=1){
                    a[idx]=x;
                    idx++;
            }
    
    
      printf("The no duplicate values are:\n");
      printf("\n%d",x);
     }
    
    
    
    return 0;
    }

  7. #7
    Epy
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    Code:
     int a[SIZE];
     int i;
     int x;
     int j;
     int flag;
     int idx=0;
    
    
      printf("Enter 20 numbers between 10 and 100\n");
      scanf("%d",&a[i]);
    &a[i] is telling scanf to place the input at the address of the 'i'th term in the 'a' array. Again, you have not supplied a value for i...so it doesn't know what part of the array you want to use. Try:

    Code:
     int a[SIZE];
     int i;
     int x;
     int j;
     int flag;
     int idx=0;
    
    
      printf("Enter 20 numbers between 10 and 100\n");
      scanf("%d",&a[0]);
    
    //this is how to accept multiple numbers
    //i = 1 because a[0] has already been filled
      for (i = 1; i < 20;i++){

  8. #8
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    I tried everything you told me, it still gives me segmentation error.

  9. #9
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    Isn't the problem that you shouldn't use the addressof operator on the array?

  10. #10
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    what do you mean?

  11. #11
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    Nevermind, I think I'm confusing things here.

  12. #12
    C++まいる!Cをこわせ! Elysia's Avatar
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    Warning 3 warning C6001: Using uninitialized memory 'i': Lines: 6, 7, 8, 9, 10, 11, 14, 15 15
    Warning 4 warning C6001: Using uninitialized memory 'j': Lines: 6, 7, 8, 9, 10, 11, 14, 15, 18, 19, 20, 23, 24, 25, 35 35
    Warning 5 warning C6001: Using uninitialized memory 'flag': Lines: 6, 7, 8, 9, 10, 11, 14, 15, 18, 19, 20, 23, 24, 25, 35, 39 39
    Warning 6 warning C4700: uninitialized local variable 'i' used 15

    What this is saying is that you are using a lot of uninitialized variables.
    Quote Originally Posted by Adak View Post
    io.h certainly IS included in some modern compilers. It is no longer part of the standard for C, but it is nevertheless, included in the very latest Pelles C versions.
    Quote Originally Posted by Salem View Post
    You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much.

    Outside of your DOS world, your header file is meaningless.

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