In C check to see if an int exists? - c

Question

I have a 13x13 array with elements but not every element is number. If its not a number, then its filled with blank space so I am trying to figure out how to check to see if a number exists at a given index. If a number exists, it does a job if not, it goes back to the loop. This is what I have:
int *spaceCheck;
int i, Value;
for(i=0;i<169;i++){
spaceCheck = &Array[i];
if(!spaceCheck){
continue;
else
//do this job
}
Can I do this? Is there any other better way to check? Thank you for you help.
Its an array of integers of 169 elements. So when the data(numbers) is initially loaded, there are only 80 elements loaded from the file into the array. The rest of the space in the array is filled with blank space (printf(" ")). Sorry for the confusion.

Answer 1

To do what you are asking, you need to use a sentinel value that indicates "no value", eg:
const int novalue = -1; // or whatever you want, as long as it is unique
int Array[169];
for (i=0;i<169;i++)
Array[i] = novalue;
...
for(i=0;i<169;i++){
Value = Array[i];
if(Value == novalue)
continue;
//do this job
}
Otherwise, create an array on structs instead of ints:
struct sValue
{
char valid;
int value;
};
sValue Array[169];
for (i=0;i<169;i++)
Array[i].valid = 0;
...
for(i=0;i<169;i++){
if(Array[i].valid == 0)
continue;
Value = Array[i].value;
//do this job
}

Answer 2

The array elements may not be initialized with zero always. The first thing you need to do is initialize it with a default value. say 0 or -1 or whatever. Then you can modify your code as follows:
int *spaceCheck;
int i, Value;
for(i=0;i<169;i++){
spaceCheck = &Array[i];
if(*spaceCheck!=default_value){
continue;
else
//do this job
}
Note that you were also storing the reference of the array element. If you want to check its data you need to use the dereferencing operator *. Hope that this is what you want. Simpler way to write the above code is to not use pointers at all. It is as follows:
int i, Value;
for(i=0;i<169;i++){
if(Array[i]!=default_value){
continue;
else
//do this job
}

Answer 3

I think here in this statement :if(!spaceCheck) you are checking the address of the variable to be true and you are not comparing it with any value.As for example ,if((*spacecheck)!=Array[i][j])//considering you have a 2-D array)will be logically correct as per your question.Well the following code may clear your concept,it is based on conditions similar to what you have provided here,
int *spaceCheck;
int i,j, Value;
for(i=0;i<13;i++){
for(j=0;j<13;j++){
spaceCheck=&array[i][j];//array is 2-D
if(*spaceCheck!=array[i][j])
continue;
else
//do this job
}}