macOS Printing int arrays like a string?

[larswik]

So a char array can be simply printed using the %s. Is their also an EASY was to print a int array for must I create some kind of loop to do it? I am doing a test writing a random number generating guessing game, and I want to populate an array with ints.

I was testing my program as I went along to check to see if the array was populated correctly. I was having to do it 1 at a time since my array was an int and not a char.

-Lars

EDIT: I found another way of doing this that works. Thanks

 

[lee1210]

There's no good terminator for an int array for the library to depend on, so no dice. Not sure what method you came up with, but a function that accepts the base of the array and its length as arguments and prints in a loop might do it. Then you don't need loops muddying up other code, you can just invoke the function.

-Lee

 

[balamw]

EDIT: I found another way of doing this that works. Thanks

Can you post more about the solution you found?

1. It might help someone else
2. Talking about it helps you grasp the concepts better
3. It may stimulate further ideas and discussion

B

 

[larswik]

Sure... The goal is to work with array's to better understand them. First I wanted to populate the char array with '-' just dashes. Then replace one of the chars with the random number that was an int, bad idea, there is a big difference between 5 and '5'. I then replaced one of the Chars with 'Y' instead of trying to do it with an int with this code: arrayHolder[randNum] = 'Y'; I have tests built in that give away the answer to see if it worked correctly.

#include <stdio.h>
#include <time.h>
#include <stdlib.h>

#define total 10

int main (int argc, const char * argv[]) {

	int randNum, i, playerGuess;
	char arrayHolder[total];
	
	srand(clock());
	randNum = rand() %total; //Gen a num 0 - 9
	
	printf("the random number is %d\n", randNum); //Test to see the num
	
	for (i=0; i<= total; i++)
	{
		arrayHolder[i] = '-';
		arrayHolder[randNum] = 'Y';
	}
	
	printf("The array looks like this: %s\n", arrayHolder);// test to look at the array
	
	printf("Enter a number from 0 to 9: ");
	scanf("%d", &playerGuess);
	
	if (arrayHolder[playerGuess] == 'Y')
		printf("You guessed it!");
	else
		printf("Wrong! it was: %d", randNum);
	
    return 0;
}

 

[lee1210]

A tip (that may not be immediately applicable):
You can subtract '0' from '0'-'9' to get the integer value for that digit.

-Lee

 

[balamw]

there is a big difference between 5 and '5'.

A tip (that may not be immediately applicable):
You can subtract '0' from '0'-'9' to get the integer value for that digit.

I think it's already on larswik's mind.

Converting from char to int in this way is a good way to learn about converting between data types and learning something about the ASCII character set. The same general technique can be useful for letters or converting uppercase to lowercase and back.

B

 

[larswik]

All right - Let me understand this since I did see / read about this a little. If I want to start with int value of lets say 2, it is stored as the ASCII value of 50 the same way that 'a' is stored as 97. ( my terminology might be a little off).

#include <stdio.h>

int main (int argc, const char * argv[]) {

	char letter;
	
	letter = 'a';
	
	printf("the letter number is %d", letter);
    return 0;
}

So if I understand you correctly, if I have char value of 9, it's stored as the ASCII of 57 and if I subtract '0' which is ASCII value 48 from ASCII 57 I get 9.

So this code would convert my char to an int. Is this called type casting or data type conversion, or do they mean the same thing?

#include <stdio.h>

int main (int argc, const char * argv[]) {

	char let1, let2;
	int result;
	
	let1 = '7';
	let2 = '0';
	
	result = let1 - let2;
	
	printf("The number value is %d", result);
	
    return 0;
}

Thanks again for you help guys!

-Lars

 

[lee1210]

All right - Let me understand this since I did see / read about this a little. If I want to start with int value of lets say 2, it is stored as the ASCII value of 50 the same way that 'a' is stored as 97. ( my terminology might be a little off).

#include <stdio.h>

int main (int argc, const char * argv[]) {

	char letter;
	
	letter = 'a';
	
	printf("the letter number is %d", letter);
    return 0;
}

So if I understand you correctly, if I have char value of 9, it's stored as the ASCII of 57 and if I subtract '0' which is ASCII value 48 from ASCII 57 I get 9.

So this code would convert my char to an int. Is this called type casting or data type conversion, or do they mean the same thing?

#include <stdio.h>

int main (int argc, const char * argv[]) {

	char let1, let2;
	int result;
	
	let1 = '7';
	let2 = '0';
	
	result = let1 - let2;
	
	printf("The number value is %d", result);
	
    return 0;
}

Thanks again for you help guys!

-Lars

Calling this "conversion" might be the most accurate here. You have a digit stored as a character using an encoding (ASCII). You want to get the integer value this character represents. So you have to "map" the encoded digit to the integer value it represents. As a convenience we use '0' to do this mapping instead of remembering it's ASCII value.

Your code looks sound to me. There is a typecast involved, but not where you think. Typecasting is taking a value of one type and getting the best match for this value as another type. This can be done explicitly or implicitly. In this case when you subtract one char from another, the result is a char. You want to store the result as an int, so the char (which is a signed integer) must be cast up to an int (which is likely a wider signed integer). This is pretty cheap because these values are stored as 2's complement (don't worry about this just yet) so the highest bit is used to "extend" the value, filling all the bits needed to make up for the difference between the width of a char and an int. A typecast can be done between all primitives, with varying degrees of difficulty/accuracy. When casting down precision is lost so bad things can happen.

-Lee

 

[Matthew Yohe]

Another example of this type of conversion, you could add 32 to any uppercase letter, and get the resulting lower case. ie. Adding 32 to 'A' and it would result in the lowercase 'a'

 

[lee1210]

Another example of this type of conversion, you could add 32 to any uppercase letter, and get the resulting lower case. ie. Adding 32 to 'A' and it would result in the lowercase 'a'

Or adding 'a'-'A'.

-Lee

 

[dmi]

char arrayHolder[total];

randNum = rand() %total; //Gen a num 0 - 9

for (i=0; i<= total; i++)
{
arrayHolder = '-';
arrayHolder[randNum] = 'Y';
}

printf("The array looks like this: %s\n", arrayHolder);// test to look at the array
}[/CODE]

i<= total should be i<total
and you should also set
arrayHolder[total]=0;
with
char arrayHolder[total+1];
to terminate the %s
arrayHolder[randNum] = 'Y'; can also be done after the loop

 

[larswik]

Cool, thanks you. I get it now!

-Lars