The value of i is huge in this for loop !?

[FlamurKasa]

Why is 'i' so big? I need it to be 1,2,3...

#include <stdio.h>
int
main()
{
     int i,n,p;
     printf("How many numbers:");
     scanf("%d",&n);
     int v[n];
     for(i=1;i<=n;i++)
{
     printf("n%d="),i;     
     scanf("%d",&v[i]);
}
     printf("N=%d\n\n",n);
     system("PAUSE");
}

[Dheeraj Joshi]

I am not aware of such syntax.

printf("n%d="),i;

Try

printf("%d\n",i);

Regards
Dheeraj Joshi

[Rabbit]

If I remember correctly, doesn't C instantiated all variables at the beginning regardless of where it was declared? That's probably what's causing it.

[FlamurKasa]

to Dheeraj Joshi:
It prints the current number(n) and its value, example:
n1=1
n2=2
...

to Rabbit:
how can I fix it???
normally the value of i should be 1,2,3,...; if you test it you get big numbers, but the programs runs perfectly if you enter 3 it loops 3 times ?!?! I am confused ?! Please test it if I'm not clear...

[tdlr]

This shouldn't even compile.

int v[n];

n needs to be a const value here and already be initialized at compile time.

Instead, this works and makes also use of the C++ style functions, which are a lot easier to use and safer:

#include <iostream>
#include <vector>

 int main()
 {
      int i,n,p;
	  std::vector<int> v;
      printf("How many numbers:");
      std::cin>>n;
      for(i=1;i<=n;i++)
		 {
			  std::cout<<i<<"=";     
			  std::cin>>p;
			  v.push_back(p);
		 }
      std::cout<<"N="<<n<<std::endl;

	  // You might now do this:
	  std::cout<<v.at(0);
	  // to print the first element
 }

Edit: Here's a c style version that works:

 
#include <stdio.h>
 int
 main()
 {
      int i,n,p; // p is unused however
      int* v;
      printf("How many numbers:");
      scanf("%d",&n);
      v = (int*) malloc(n*sizeof(int));
      for(i=0;i<n;i++)
		 {
			  printf("n%d=",i);     
			  scanf("%d",&v[i]);
		 }
      printf("N=%d\n\n",n);
      free(v);  // Don't forget this
 }

[donbock]

In what way is i huge?

[FlamurKasa]

to donbock:
the value printed by printf("%d",i) which should be 1,2,3..
is instead 528465523,25846 54,21576854,... .

[donbock]

Have you followed the advice given by Dheeraj? I think that would have fixed your problem.
Line 11 of your program is:

     printf("n%d="),i;

Notice that ,i is outside the parentheses. This means that i is not passed to printf. Instead, printf prints whatever garbage happens to be on the stack. When printf returns the compiler runs into the comma operator, which is effectively a semicolon. Then the compiler finds a statement consisting of a bare i; this is effectively a null statement (nop). Perhaps you are used to cout (in C++); printf works differently.

Another issue is that this printf is inside the loop and it doesn't emit a newline. This means that if the code were otherwise working and n were 5, the output would be "n1=n2=n3=n4=n5 =" -- the concatenation of all the i values.

If you really want the output to be
n1=1
n2=2
...
then you want

     printf("n%d=%d\n",i,i);

but I don't know why you want to print the value of i twice.

By the way, when you declare an array as v[n] then the array index must be between 0 and n-1. Your index ranges between 1 and n. The last array access goes off the end of the array -- that's undefined behavior.

[Dheeraj Joshi]

Hi, tdlr

v = new int[n];

In you C version of the code you have mentioned new operator. There is no new operator in C.

Regards
Dheeraj Joshi

[tdlr]

@Dheeraj Joshi:
Of course, you're right. And of course I meant:

v = (int*) malloc(n*sizeof(int));
/*...*/
free(v)

[FlamurKasa]

@donbock
Finally a solution!(sorry for my late response)
You found my mistake thank you very much, and thanks everyone else for trying.

As for your other advices, they are not so true:
you state that <the output would be "n1=n2=n3=n4=n5 =">
let me point to the next line - scanf
after each value given the program goes automatically to a new line.
The purpose of the loop was store numbers from the user into v[n], so I don't need to print the value of i 2 times.
I just want the user to know at which number he is (n2,n6, n99...). I think you got the point.

And yes i in v[i] must start at zero, I knew it but I was rushing.

The final version of the complete code:

// Shows the numbers appearing more than once in a 
//given array
#include <stdio.h>
int
main()
{
    int i,j,n,t=0,p=0;
    printf("How many numbers do you have:");
    scanf("%d",&n);
    int v[n];
    for(i=0;i<n;i++)
{
    printf("\n n %d = ",i+1);
    scanf("%d",&v[i]);
}

    for(i=0;i<n-1;i++)
       for(j=i+1;j<=n;j++)
       if (v[i]>v[j])
       {
       t=v[i];
       v[i]=v[j];
       v[j]=t;
       }       
 printf("\n This numbers occurs more than once: ");
 for(i=1;i<n;i++)
 {
if(v[i]==v[i-1])
if(v[i]!=v[i+1])
{
    printf(" %d",v[i]);
    p++;
}
}
if(p==0)
printf(" every number is unique.");
    printf("\n\n");
    system("PAUSE");
}

Tell me what you think!

[donbock]

Line 10. FYI, a C89 compiler would flag this line as an error. That version of the language doesn't allow declarations to appear below executable statements. That version of the compiler doesn't support run-time specification of array size. You might want to add a comment to announce this limitation. Better would be to use conditional compilation to execute a #error if compiled with the wrong compiler version.

Lines 13-14. I realize now that you expect the scanf characters to be echoed to stdout. I'm not familiar enough with scanf to know if that is a portable assumption. Perhaps one of the other experts can tell us.

Lines 18-23. j will count up to n. In that last iteration the references to v[j] go beyond the end of the array.

Line 29. When i==n-1, v[i+1] goes beyond the end of the array.

[FlamurKasa]

18.  for(j=i+1;j<=n;j++)

26.  for(i=1;i<n;i++)
        if(i!=(n-1))
{
            if((v[i]==v[i-1])&&(v[i]!=v[i+1]))
    {
    printf(" %d",v[i]);
    p++;
    }
}
        else if(v[i]==v[i-1])
        printf(" %d",v[i]);

should be fine now (scanf is like cin>> in c++)