This question already has answers here:
Closed 10 years ago.
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Macro definition error in C?
I'm new to programming and hope someone can help me with this:
Why is it giving an output : 5 .
Here is the code snippet:
#include <stdio.h>
#define max 5;
int main(){
int i=0;
i = max+1;
printf("\n%d",i);
return 0;
}
Because the macro has a semi-colon. Code is equivalent to:
i = 5; + 1;
Remove the semi-colon from the macro.
The crucial point is
#define max 5;
When substituted by the preprocessor, this will become
i = 5; +1;
which assigns 5 to the variable i (the expression +1; has no effect).
You need to write
#define max 5
Because you have a semicolon after your macro definition x=
Related
This question already has answers here:
The need for parentheses in macros in C [duplicate]
(8 answers)
Closed 2 years ago.
My code:
#include <stdio.h>
#define PRODUCT(x) (x * x)
int main()
{
int i = 3, j, k, l;
j = PRODUCT(i + 1);
k = PRODUCT(i++);
l = PRODUCT(++i);
printf("%d %d %d %d", i, j, k, l);
return 0;
}
I am not able to comprehend why the output is:
7 7 12 49.
Is there any error in macro or some other problem?
Your code has undefined behavior, operations in i:
k=PRODUCT(i++);
l=PRODUCT(++i);
lack a sequence point.
As for:
j=PRODUCT(i+1);
It expands to i+1*i+1 which is i+i+1 which is 7. I assume it's not the expected result, in the future also include that in your question.
Your macro is incorrect. The following expression:
PRODUCT(i+1)
will expand to
(i+1 * i+1)
which is 2*i+1.
Your macro should be:
#define PRODUCT(x) ((x)*(x))
I strongly suggest you stop using macros for this sort of thing. You could easily write this as a function:
int product(int x)
{
return x * x;
}
Note that this will only work for the example I gave. If you try
PRODUCT(i++)
you will get
( (i++) * (i++) )
which invokes undefined behaviour, as this expression lacks a sequence point between the 2 increments.
This question already has answers here:
What is argument evaluation?
(2 answers)
Closed 6 years ago.
Problems using define in C. Works well when I call OP(9), but when I call OP(7+2) I get 23. Why?
#include<stdio.h>
#include<stdlib.h>
#define OP(x) x*x;
int main() {
int x,y;
x = 2;
y = OP(7+2);
printf("%d", y);
return 0;
}
Why prints 23 and not 81?
You should wrap x in parentheses in order to force precedence. However, it is also essential to wrap the entire expression in parentheses.
#define OP(x) ((x)*(x))
This question already has answers here:
Syntax error while using a #define in initializing an array, and as arguments to a function in C? [closed]
(2 answers)
Closed 5 years ago.
I've tried to figure out this behavior of the define command in C (I'm new with that). I've got this code and I don't know why I see in the output that myAge=15 and not 16 (I know it's 15, but I don't know why). Anybody can help me to find out why does it happen?
this is the code:
#include <stdio.h>
#include <stdlib.h>
#define AGE 15;
int main(void)
{
float myAge = AGE + 1;
printf("Hello!\n");
printf("My name is Raz and I am %d years old!\n", myAge);
system("PAUSE");
return 0;
}
Thanks for helping :)
#define is a textual replacement performed by the preprocessor prior to the compilation step. In this case, you're asking the preprocessor to expand the token AGE to 15;. The semicolon is part of the expansion, so this is the code you would get after the preprocessing step:
float myAge = 15; + 1;
As you can see, it does not expand to what you expect.
You can fix this issue by removing the semicolon from the #define:
#define AGE 15
Better yet, avoid using the preprocessor for simple numerical constants - consider using a const int instead:
const int age = 15;
This question already has answers here:
Strange behaviour of macros C/C++
(5 answers)
Closed 8 years ago.
I have the following piece of code.
#include <stdio.h>
#define d 10+10
int main()
{
printf("%d",d*d);
return 0;
}
As 10+10=20, I thought that d would be 20 everywhere in the program. But when I execute d*d, I have expected the result to be d*d=20*20=400. But the result gets printed as 120. Can anyone give me an explanation for this behavior?
Preprocessor is just doing simple "find & replace", so this code:
printf("%d",d*d);
changes to
printf("%d",10+10*10+10);
which is 10+100+10 = 120
That's why it's so important to add parens in defines:
#define d (10+10)
printf("%d",d*d);
Here d is replaced as printf("%d",10+10*10+10). so in this case it first executes 10*10 and adds with 10+100+10. so results 120.
To eliminate this-
#define d (10+10) // FIX
int main()
{
printf("%d",d*d);
return 0;
}
Change
#define d 10+10
to
#define d (10+10)
What you get without the proper parentheses is 10+10*10+10 which is the same as 10 + (10 * 10) + 10
printf("%d",d*d) will evaluate (after preprocessing) to printf("%d", 10+10*10+10);
This question already has answers here:
Printing array elements
(6 answers)
Closed 9 years ago.
The expected output of the following C program is to print the elements in the array. But when actually run, it doesn't do so.
#include<stdio.h>
#define TOTAL_ELEMENTS (sizeof(array) / sizeof(array[0]))
int array[] = {23,34,12,17,204,99,16};
int main()
{
int d;
for(d=-1;d <= (TOTAL_ELEMENTS-2);d++)
printf("%d\n",array[d+1]);//printing the array
return 0;
}//looks simple but no result
What's going wrong? Why am I not getting any output?
In the comparison
d <= (TOTAL_ELEMENTS-2)
TOTAL_ELEMENTS has type size_t so d is converted to unsigned. For, say, sizeof(size_t)==4, this makes the test
0xffffffff < 5
which fails, causing the loop to exit.
If you really want to start your loop counter from -1
d <= (int)(TOTAL_ELEMENTS-2)
would work