Float & Double

Questions about float and double is relatively difficult, but you may be asked if the position you applied for is related to DSP programming.

Double Precision Format

Exponent	Equation
0x000	(−1)^signbit × 2^−1022 × 0.significandbits
0x001, …, 0x7FE	(−1)^signbit × 2^(exponentbits−1023) × 1.significandbits
0x7FF	Not Define(infinite)

Example:
3ff0 0000 0000 0000 = 1 Note: 1.xxx in decimal, must has 3ff… in binary
3ff0 0000 0000 0001 = 1.0000000000000002, the next higher number > 1
3ff0 0000 0000 0002 = 1.0000000000000004
4000 0000 0000 0000 = 2
c000 0000 0000 0000 = −2

int main()
{
    float a = 1.1f;
    if (a == 1.1) 
        printf("a is equal to 1.1");
    else 
        printf("a is not equal to 1.1");
    return 0;
}

Ans:
Output "a is not equal to 1.1", because constant 1.1 will be stored in double precision format (default), while a is stored in float.

Print It Out



int main()
{
    int i, zero;
    long long temp1;
    long temp2;
    double d = -1.25;
    float f = -1.25f;
 
    printf("-1.25 double:   ");
    temp1 = *((long long*)&d);
    zero = (temp1 == 0ULL)?1:0;                // used to determine default bit
    for (i = 0; i < 64; i++) {
        if (i == 1 || i == 12)
            printf(" ");
        if (i == 12)
            printf("(%d).", !zero);            // always (1). except for 0.0 case
        printf("%d", temp1 < 0);               // double can't &
        temp1 <<= 1;                           // use the sign bit                     
    }
    printf("\n");
 
    printf("-1.25f float:   ");
    temp2 = *((long*)&f);
    zero = (temp2 == 0UL)?1:0;
    for (i = 0; i < 32; i++) {
        if (i == 1 || i == 9)
            printf(" ");
        if (i == 9)
            printf("(%d).", !zero);
        printf("%d", temp2 < 0);
        temp2 <<= 1;
    }
    printf("\n");
}

Output:
-1.25 double: 1 01111111111 (1).0100000000000000000000000000000000000000000000000000
-1.25f float: 1 01111111 (1).01000000000000000000000