Expected 'char' but argument is of type 'char *' what is the problem ?

Question

0.00/5 (No votes)

See more:

C++

void display(char, char*, float, int, float);

display(name,item,price,qtty,payment);

void display(char name,char*item,float price,int qtty,float payment)
{
	fflush(stdin);
	printf("\n----------------------------------------------------");
	printf("\n-                RECEIPT OF PURCHASE               -");
	printf("\n----------------------------------------------------");
	printf("\nName        : %s",name);
	printf("\nItem        : %s",item);
	printf("\nPrice       : RM %.2f",price);
	printf("\nQuantity    : %d",qtty);
	printf("\nPayment     : RM %.2f",payment);
	printf("\n----------------------------------------------------");
}

What I have tried:

The display output just only

----------------------------------------------------
                 RECEIPT OF PURCHASE
---------------------------------------------------

Posted 15-Jan-21 7:25am

tan alan

Updated 15-Jan-21 23:05pm

Richard MacCutchan

v3

Add a Solution

1 solution

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

OriginalGriff · Answer 1 · 2021-01-15T08:03:00

Solution 1

When you call printf, you can provide a number of arguments. The first is the format, and it is mandatory - it tells the function what it is to do.
The format string you provide contains characters to print "\nName: " and "type specifiers" which describe the rest of the arguments - these are always a percent character followed by at least one other character, which describes the data.

%c	character
%d	decimal (integer) number (base 10)
%e	exponential floating-point number
%f	floating-point number
%i	integer (base 10)
%o	octal number (base 8)
%s	a string of characters
%u	unsigned decimal (integer) number
%x	number in hexadecimal (base 16)
%%	print a percent sign

You are telling it to print a string:

printf("\nName : %s",name);

But name is defined as a character:

void display(char name, ...

A string requires an address, not a character value!

I'd suspect that you want to pass a char* to your display function, since most people have more that a single character in their names!

Posted 15-Jan-21 8:03am

OriginalGriff

Comments

tan alan 15-Jan-21 14:14pm

i change the name to the username as variable,but the display display did not come out the print out function

OriginalGriff 15-Jan-21 14:28pm

And what did you pass? I can't see your code, remember?

tan alan 15-Jan-21 14:40pm

i change to printf("\nName : %c",name); its display the equal sign symbol what it means ?

OriginalGriff 15-Jan-21 14:52pm

And what did you pass to the "display" function when you called it?

tan alan 16-Jan-21 0:37am

same as
void display(char name,char*item,float price,int qtty,float payment)

OriginalGriff 16-Jan-21 3:31am

:sigh:

That's not passing a value to the method, that's *declaring( the function and it's parameters.

Have a look at the code that *calls* the function ...

tan alan 16-Jan-21 5:27am

so what should i correct ?@@

OriginalGriff 16-Jan-21 5:57am

How should I know? I can't see what you did - which might just be important, if you think about it ...

OriginalGriff 16-Jan-21 6:33am

Banging your head on a desk is wonderful.
It feels so good when you stop ...

And right at the moment, I'm wearing a hole in mine...

Look at your code.
Find everywhere you call display.
Then look at each call, and find out exactly what you are passing to it with that call - the variable name.
Then look at your code to find out what that variable is declared as.
Then look at your code again, and find out what type of parameter display expects.

What do you see?

tan alan 16-Jan-21 6:36am

i did not found any error..see from my side...

OriginalGriff 16-Jan-21 7:04am

:sigh:
"Find everywhere you call display.
Then look at each call, and find out exactly what you are passing to it with that call - the variable name."
display(zname,item,price,qtty,payment);
(twice)

That'll be "zname" you are passing then.

"Then look at your code to find out what that variable is declared as."
char zname[10],item[40];

So, that's an array of characters then.

"Then look at your code again, and find out what type of parameter display expects."
void display(char zname, char* get_item, float, int, float);
And
void display(char zname,char *item,float price,int qtty,float payment)
{
...
}

So that's be a single character then.

What part of that was difficult for you to follow and do yourself?
And what part of what I originally wrote 17 hours ago did you actually read?

tan alan 16-Jan-21 8:07am

The problem I expected is the zname their wrong but i dont know which is the problem

OriginalGriff 16-Jan-21 8:36am

Oh come on!
Forget names. As far as the compiler is concerned they are irrelevant - you *know* that!

Stop panicking, start thinking.
What did I say just over an hour ago?
Does anything there look inconsistent?
(Hint: yes.)
If so, what?