How to print an index of an int array

Question

I have an array of ints, denoted as int* myInts[100];

When I try to print the 0^th index like so,

printf("%d\n",myInts[0]);

I get the following compiler warning:

warning: format ‘%d’ expects argument of type ‘int’, but argument 2 has type ‘int *’ [-Wformat=]

Why is this?

Mohit Jain · Accepted Answer · 2014-09-22 04:47:39Z

4

If you want an int array, declare it as

int myInts[100];

If you want array of int *, print it as

printf("%p\n", (void *)myInts[0]);

answered Sep 22, 2014 at 4:47

Mohit Jain

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3 Comments

Mohit Jain Over a year ago

int * means pointer to int and [100] means array of size 100. Collectively an array of size 100 containing int pointers.

Apollo Over a year ago

how would I initialize all of the indexes in the array to 0?

Mohit Jain Over a year ago

int myInts[100] = {0};

paxdiablo · Accepted Answer · 2014-09-22 04:49:59Z

1

int* myInts[100];

declares an array of one hundred pointers to integers, not one hundred integers as you describe.

You need to use:

int myInts[100];

instead.

answered Sep 22, 2014 at 4:49

paxdiablo

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Comments

Anbu.Sankar · Accepted Answer · 2014-09-22 07:07:34Z

1

int* myInts[100]; It means array of 100 integer pointer. int myInts[100]; It means array of 100 integer.

to eliminate warning do like printf("%p\n", (void *)myInts[0]);

edited Sep 22, 2014 at 7:07

answered Sep 22, 2014 at 4:52

Anbu.Sankar

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1 Comment

mch Over a year ago

you should remove the * in the second declaration to match the text.

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