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Start your free trialShanmukh Karra
678 PointsProblems with Variables - XCode
Hello There! I have tried the "char" variable in XCode as part of Stage 1 of Objective C Development. Upon entering the variable, I am always greeted with an error that states, "Incompatible pointer to integer conversion assigning to 'char' from 'char [2]".
My code reads:
int main( )
{
int days_in_a_week = 7;
float cm_to_in = 2.54;
char the_w = "W";
printf("%d days in a week.\n", days_in_a_week);
printf("%f cm per in.\n", cm_to_in);
printf("The %c is a cool hotel.\n", the_w);
return 0;
}
The output reads:
7 days in a week. 2.540000 cm per in. The 4 is a cool hotel. Program ended with exit code: 0
Please help with the error and why "W" is output at "4".
3 Answers
Stone Preston
42,016 Pointstry using single quotes around your char. chars use single quotes, whereas strings use double quotes. what happens when you use double quotes is that it thinks you want to make an array of characters, which explains the error message
int main()
{
int days_in_a_week = 7;
float cm_to_in = 2.54;
char the_w = 'W';
printf("%d days in a week.\n", days_in_a_week);
printf("%f cm per in.\n", cm_to_in);
printf("The %c is a cool hotel.\n", the_w);
return 0;
}
Shanmukh Karra
678 PointsAlso, when I had put up the code in my question, why was the code beginning only from "char the_w = "W";" shown? Was it one of those formatting errors?
Stone Preston
42,016 Pointsyes. the below animation shows how to post a code block. you have to skip a line after any previous text if you want it to be shown correctly. Ill edit your original post for you so you can see what I mean. just click edit and you can see my additions
Shanmukh Karra
678 PointsThank you so much! I had done it myself. :D
Shanmukh Karra
678 PointsShanmukh Karra
678 PointsOh yes... no wonder it was not working. Just a stupid beginner's question. Thanks for helping out so quickly.