Characters in if statements

My goal here is to define 'target' yet when running the program, entering either m or f yields an undefined 'target'. Can anyone identify my problem?

Quote:

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int age, gender, num_rates = -1, hit_rates = 0, rate, target;

printf("What is your gender? (m or f)\n");
scanf("%c", &gender);
printf("What is your age?\n");
scanf("%d", &age);
printf("Enter your recorded heart rates:\n");
if (gender == 'f')
target = 220-age;
else if (gender == 'm')
target = 226-age;

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Thanks

You are declaring "gender" as an integer, but assigning a character to it with "scanf()."

Quote:

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Originally Posted by Matticus

You are declaring "gender" as an integer, but assigning a character to it with "scanf()."

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I dont think that would be problem. There would be implicit coversion.
@OP: you dont read the heart rate, have you noticed that. And what is the expected out for the 'm' and age 25? I dont see any issues.

ssharish

Quote:

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Originally Posted by ssharish2005

I dont think that would be problem. There would be implicit coversion.
@OP: you dont read the heart rate, have you noticed that. And what is the expected out for the 'm' and age 25? I dont see any issues.

ssharish

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Actually, that was the problem, and I only posted a portion of the entire program. The entire thing is functional now. Thanks, Matticus.

Nigel... you really should check the return values of scanf() to be sure the conversions succeeded... It returns the number of successful conversions so in your case anything but 1 is an error.

No problem.

@ ssharish, my compiler doesn't like this, either.

Code:

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#include <stdio.h>

int main(void)
{
        int x;

        printf("Enter value for 'x':  ");
        scanf("%c",&x);

        printf("%d\t%c\n\n",x,x);

        if(x=='m') printf("This won't print.\n\n");

        return 0;
}

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Quote:

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Originally Posted by Matticus

No problem.

@ ssharish, my compiler doesn't like this, either.

Code:

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#include <stdio.h>

int main(void)
{
        int x;

        printf("Enter value for 'x':  ");
        scanf("%c",&x);

        printf("%d\t%c\n\n",x,x);

        if(x=='m') printf("This won't print.\n\n");

        return 0;
}

---

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What compiler do you use? have no issue compiler and running it. See my output

Code:

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Enter value for 'x':  m
109    m

This won't print.

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ssharish

Quote:

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Originally Posted by ssharish2005

What compiler do you use? have no issue compiler and running it. See my output

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The real issue is what compiler do YOU use?

Since you are assigning a char to an int the conversion in scanf() could fail.


From Pelles C...

Quote:

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Building main.obj.
E:\C_Code\Experiments\testprog\main.c(8): warning #2234: Argument 2 to 'scanf' does not match the format string; expected 'char *' but found 'int *'.
Building testprog.exe.
Done.

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It has to do with native machine's byte order. When scanf assigns just one byte into a 4-byte integer, the remaining bytes may still contain garbage. The assigned byte may be placed in the lowest order byte or highest. Then when you attempt to do a comparison with 'a', as long as the other bytes aren't zero or the expected byte is in any place other than the lowest, it will always fail.

Just change x to be of char type, and it will work.