Why doesn't the second for loop execute?

Question

printf debugging. The while loop executes, the 1st for loop executes, the 2nd for loop does not.

while (size > 0){
    printf("in the while loop!\n");
    VertexPointer current = graph[0];
    int i;
    for (i = 1; i < size; i++){
        float dist = 0;
        printf("in the first for loop for i = %d \n", i);
        int j;
        for(j = 0; j++; j < dimension){
            printf("dist loop: %d\n", j);
            printf("current: %f\n", (*current).loc[0]);
            printf("unvisited: %f\n", (*graph[j]).loc[0]);

            dist = dist + pow((*current).loc[0] + (*graph[j]).loc[0], 2);
            printf("distance:  %f", dist);


            dist = sqrt(dist);
            if (dist < (*graph[i]).key){
                decreaseKey(graph, i, dist);
            }
        }

    }

    extractMin(graph, size);
    size = size - 1;
    mst_length = mst_length + (*current).key;
}

You switched the condition and increment: j++; j < dimension should be: j < dimension; j++ — Kninnug
– Kninnug, Commented Mar 7, 2014 at 22:42

Kninnug · Accepted Answer · 2014-03-07 23:08:15Z

2

You switched the condition and increment:

for(j = 0; j++; j < dimension)

so the loop tests the value of j++, and since the postfix ++ operator returns the previous value of the variable it increments, it will return 0 (the initial value of j) on the first iteration and thus never loop. Change it to

for(j = 0; j < dimension; j++)

If you turn on warnings you should receive a message that the third expression has no side-effects (and would be optimized-away by the compiler).

Alternatively (stop: don't actually do this):

change the condition to ++j to make it a (potentially) never-ending loop instead:

for(j = 0; ++j; j < dimension)

if you're lucky (which is usually, in this case) the loop will end when j reaches MAX_INT and wraps around to 0. But the standard doesn't guarantee that, so it's potentially endless.

edited Mar 7, 2014 at 23:08

answered Mar 7, 2014 at 22:45

Kninnug

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2 Comments

Filipe Gonçalves Over a year ago

Aside from the result of integer overflow being undefined by the standard (in contrary to unsigned overflow), still, in most implementations, it will wrap from INT_MAX to INT_MIN, not to 0.

Kninnug Over a year ago

Which is why I said you'll usually be lucky, but no guarantees.

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