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int a[10];
    int i;
    for (i=0; i<10; i++)
      a[i] = i;
    printf("%d", a[-1]);
    return 0;
Posted 11-Feb-13 4:34am
Edited 11-Feb-13 4:42am
Maciej Los151.8K
v2
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Solution 1

0. You are already accessing the array at -1 position which is outside the array.
1. Don't do this.
2. Whatever the reason for wanting to do this there is a better way to do it.
3. See number 1.
...
...
-1. BANG! your program is dead.
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Comments
Maciej Los at 11-Feb-13 9:43am
   
Short and to the point! +5!
Albert Holguin at 11-Feb-13 11:24am
   
Pretty much... +5
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Solution 2

You can't, if you mean the last but one element then you have to do a[(sizeof(a) / sizeof(a[0])-1].
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Solution 3

Yes, you can access to an array element [i-1] only if the index of position is bigger then 0 (in zero-based arrays).
 
    int i = 0;
    int j = 0;
    for (i=0; i<10; i++)
    {
        printf("%d", a[i]);
        if (i>0)
        {
        j=i-1;
        printf("%d", a[j]);
        }
    }
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This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

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0 Marcin Kozub 330
1 OriginalGriff 256
2 Sergey Alexandrovich Kryukov 215
3 Praneet Nadkar 197
4 Richard MacCutchan 182
0 OriginalGriff 8,048
1 Sergey Alexandrovich Kryukov 7,287
2 DamithSL 5,614
3 Manas Bhardwaj 4,986
4 Maciej Los 4,910


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