How to get index of structure elements?

Hi
I want to get the starting index of structure elements, whoz id are 0,1,2,3
Like in below code

col_data[0] (starting id=0)
col_data[3] (starting id=1)
col_data[5] (starting id=2)
col_data[8] (starting id=3)

Code:

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Code:

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typedef struct gg{
  int id;
  int value;
} gg;

gg col_data[10]={{0,44},{0,46},{0,39},{1,25},{1,10},{2,46},{2,78},{2,55},{3,64},{3,79}};
int ptr[4];

for (i = 0; i < 4; i++){
for (j = 0; j < 10; j++){
if (col_data[j].id==i){
ptr[i]=j;
// If i use break it will break after first match and will not go again to the loop. 
break;
    }
  }
}

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output should be
ptr[0]=0
ptr[1]=3
ptr[2]=5
ptr[3]=8

How can i skip remaining loop iterations when it get that index and will go back to loop again for getting next element index?

Code:

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for (i =0, j = 0; i < 4 && j < 10; ++j)
{
    if (col_data[j].id == i)
    {
        ptr[i++] = j;
    }
}

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It won't work.. continue looping and the result was wrong too

Quote:

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Originally Posted by DRK

Code:

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for (i =0, j = 0; i < 4 && j < 10; ++j)
{
    if (col_data[j].id == i)
    {
        ptr[i++] = j;
    }
}

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I think it is because i was trying this with 0,2,1,3

Code:

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// What is the efficient way of doing this in this case?

gg col_data[10]={{0,44},{0,46},{0,39},{2,25},{2,10},{1,46},{1,78},{1,55},{3,64},{3,79}};

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Quote:

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Originally Posted by gevni

I think it is because i was trying this with 0,2,1,3

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Sorting col_data by id would be cool.
Otherwise, searching for indexes in a meshed up col_data will lead to two nested loops.

//Edit

Thinking again, would you mind giving us more info about the order of elements ?
Are ids distinguished and contiguous ? If so, then loop once over the table and store each unique reference to a helper table.

Quote:

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Originally Posted by gevni

It won't work.. continue looping and the result was wrong too

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No, it won't continue. i < 4 condition terminates loop when all 4 indexes are found.

Here's a new solution for you:

Code:

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for (i = 0; i < 4; ++i)
{
    ptr[i] = -1; // index not yet found
}

for (i = 0, j = 0; i < 4 && j < 10; ++j)
{
    int id = col_data[j].id;
    if (id < 4 && ptr[id] == -1)
    {
        ptr[id] = j;
        ++i;
    }
}

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Thanks for your help. It work fine if col_data[10] is in below order

Code:

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gg col_data[10]={{0,44},{0,46},{0,39},{1,25},{1,10},{2,46},{2,78},{2,55},{3,64},{3,79}};

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I am sorry i provide the data into wrong order it should be like that

Code:

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gg col_data[10]={{3,44},{3,46},{2,39},{2,25},{2,10},{1,46},{1,78},{0,55},{0,64},{0,79}};

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In that case if i run the above code it shows me output as

Code:

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ptr[0]:7
ptr[1]:0
ptr[2]:0
ptr[3]:0

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As it find the first element in the last of array. How can i copup with it?

Quote:

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Originally Posted by DRK

No, it won't continue. i < 4 condition terminates loop when all 4 indexes are found.

Here's a new solution for you:

Code:

---

for (i = 0; i < 4; ++i)
{
    ptr[i] = -1; // index not yet found
}

for (i = 0, j = 0; i < 4 && j < 10; ++j)
{
    int id = col_data[j].id;
    if (id < 4 && ptr[id] == -1)
    {
        ptr[id] = j;
        ++i;
    }
}

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Thank you for your help.. It works :)