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Details of C that every noob forgets


I am still forgetting some of them, that’s why I wrote this post ;)

C Details

Floating numbers

The IEEE 754 standard specifies that a floating-point number will be represented as follows:

  • Sign bit: 1 bit
  • Exponent width: 8 bits
  • Significant precision: 23 bits

Read more here

And a double-precision floating-point number:

  • Sign bit: 1 bit
  • Exponent width: 11 bits
  • Significant precision: 52 bits

The % operator cannot be applied to a float or double.

Parenthesis counts

!= has a higher “range” or precedence than “=”, which means that:

c = getchar() != EOF

is equivalent to

c = (getchar() != EOF)

*p++ is not the same as (*p)++. In the first case p++ is performed first. Read more about C Operator Precedence.

In general, use parenthesis and/or split statements when they could be confusing such as:

int a, b=2, c=3;
a=b=c;
printf("a=%d\tb=%d\tc=%d\n",a,b,c);

/* Output */
a=3	b=3	c=3

It is better write that code as:

int a, b=2, c=3;
b=c;
a=b;

Be careful with the quotes

  • 'a' is not the same as "a". 'a' is a char with the numeric value of 97 (0x61), whereas "a" is an char array defined as [ 'a', '\0']. Where '\0' is the null character.

Please, give me a soul

  • extern and static variables are initialized to zero by default. Automatic and register variables will have an undefined values if they are not initialized.

  • malloc() does not initialize the allocated memory, calloc() does.

Assignment operators

Do not forget that:
x *= y+a is the same as x = x * ( y+a )

C gives a shit about indentation

In case of possible ambiguity it is better to use braces to avoid bugs.

/* Clear case:*/
if ( n>9 )
    n++;
else
    n--;

In the code above, the compiler will have no problem in understand what you want to do, but is the following case, there will be same problems even though it is “clear” for a human to understand what you want to do. It seems that else belongs to the outer if but, in reality, it does not.

/* Case with ambiguity*/
if ( n>9 )
    for ( i=0 ; i<3; i++ )
        if ( i*4 > n )
            i++;
else
    n--;

Thus, use braces as follows:

/* Case with ambiguity*/
if ( n>9 ){
    for ( i=0 ; i<3; i++ )
        if ( i*4 > n )
            i++;
    }
else
    n--;

My mind is here but my body somewhere else

Do not forget the keyword extern when you are going to use a variable that is it defined in some other file.

/* file 1*/
extern int f[];

void main(int x){
    f[0] = x;
}

/* file 2 */
int f[] = {4,5,6};

Of course, do not forget to link that file to your source. For example gcc -o a.out file1.c file2.c.

Breaks

DO NOT FORGET that break statement does NOT provide an early exit from if, but it does from for, while, do and switch.

Strings and commas

Be careful with the commas when defining char arrays, or working with strings in general.

char *words[] = {
    "hi",
     /*no comma*/
    "hello"
    "danke",
    "hola"
}

If we print each element of that array, we will get this:

    "hi"
    "hellodanke"
    "hola"

Since C concatenates strings the program printed "hellodanke".

Recommendations

Use keyword to keep your code safe

  • Cast signed or unsigned if the operand will be have either positive and negative values, or non-negative values. So the compiler doesn’t have to choose.
int i =-1;
unsigned int j = 2;

/* don't forget that 
 i = 0xFFFF
 j = 0x2   */
if ( i < j)  
    puts("i < j");
else 
    puts("actually, i > j");

The above code prints "actually, i > j, since j is unsigned so i is also treated as unsigned. Thus, 0xffff > 0x2. You should cast j as signed to make the condition true.

if ( i < (signed) j)  
    puts("i < j");
else 
    puts("actually, i > j");
  • Use const if you don’t want the variable to change (read-only status), or it shouldn’t be modified in a function.
  • Use static if you want to hide variables or functions declarations inside modules. It is similar to private keyword in c++;

Use void due to backwards compatibility

void states explicitly as empty list. Read more here

The code below works but there is no verification. test is not supposed to return anything, but due to backwards compatibility test could be defined as if it returned something. And this my friend is a bug.

/*  old C style  */
/* declaration */
test( int ); 

int main(){
    test(5);
    return 0;
}

/* definition */
test( input1 )
int input1;
{
    return (input1 + 4);
}

Last words

That’s all I have in mind for this moment. I hope this post helped you as much as it helped me. I will update this post if I remember more mistakes or if you send me more.

Cheers :3


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