A simple C question

[b_girl]

A friend of mine is just learning how to code in C and I've been helping her a little bit along the way. There seems to be a problem with a chunk of her code, and I can't figure out how to fix it, it's been quite a while since I've looked at C so I'm a little 'fuzzy' on it. The following bit of code is taken right from her source code.

int *suit[4];
for(int p = 0; p <= 3; p++)
{
suit[p] = &p;
}

... then inside a loop which will execute the following loop x = 0 to x < 4 times: printf(" %d", *suit[x]); The program exits fine, with no errors, but the only problem is, when it prints out at the end, each entry in the suit array is set at being 4 (so the 4 will print out 4 times), when what she want to print out is 0 1 2 3, if she changes the loop to: for(int p = 0; p <= 4; p++) { ... } the print statement shows that a 5 prints out in each case. I'm sure the problem comes from this suit[p] = &p; but I don't know how to fix it. Can anyone help us out? Thanks...

[markkuk]

There's only one variable p, and every element of the array suit is made to point to that element. What exactly are you trying to do in this code?

[John M Drescher]

Generally in a loop c programmers will do this (remove the =):

int *suit[4];
for(int p = 0; p < 4; p++)

This will run through all 4 suits (0..3). John

[noahsarf]

essentially what you are doing is setting each of your array elements to the address of p; At the end of your loop you have an entire array of addresses to the p value. This is why you print out a bunch of 4's or 5's depending on your loop criteria. If you are just trying to set the array locations to the value of the integers, then there is no need to use address. create your array as : int suit[4]; this is an array of integers. Now inside the for loop, suit[p]=p; will now set your array entry to the integer value of p and you will receive your expected results from your printf statement.

[b_girl]

What we would like to have happen is to have suit[0] = 0 and suit[1] = 1, etc... but suit[] is a pointer, int *suit[4]; So how, in that for loop, do we say that we want suit[p] = p itself, and not the address of p? we also tried: *suit[p] = p; and suit[p] = p; neither of these works.

[b_girl]

noahsarf wrote: Now inside the for loop, suit[p]=p; will now set your array entry to the integer value of p doing it that way gives the error: '=' cannot convert from int to int* that was one of the original ways we tried it.

[Jesse Evans]

As noahsarf said, you need to set each member of suit[] to the value of p, thus: suit[p]=p; the reason why this did not seem to work for you is that you need to also change your print statement; instead of printf(" %d", *suit[x]); you need to say printf(" %d", suit[x]); Good Luck!! 'til next we type... HAVE FUN!! -- Jesse

[Rick York]

Remember to declare suit as : int suit[4]; The Ten Commandments For C Programmers

[b_girl]

Thank you so much! As it turns out, she has no need to use pointers anyway, so we've gotten rid of all the pointers that she's been using.

[John R Shaw]

This is C++:

int *suit[4];
for(int p = 0; p <= 3; p++)
{
suit[p] = &p;
}

This is C:

int *suit[4];
int p;
for(p = 0; p <= 3; p++)
{
suit[p] = &p;
}

This is what I think your are trying to do since what you are doing makes no since:

int suit[4];
int p;
for(p = 0; p <= 3; p++)
{
suit[p] = p;
}

One last note: Get out of the habit for writing p++ when not neccessary, use ++p instead. When your friend graduates from C to C++ it will eventualy save her a lot of head acks, because post increment can lead to a great deal of unexpected over head when using user define data types. INTP