I am adding two values together and it is rounding up but I don't need it to round up.

material_cost = total * 3.67;

shipping_cost = total * 0.73;

total_cost = material_cost + shipping_cost;

fprintf('The material cost is $%0.2f. \n', material_cost);

fprintf('The shipping cost is $%0.2f. \n', shipping_cost);

fprintf('The total cost is $%0.2f. \n', total_cost);

so it prints out:

The material cost is $19.76.

The shipping cost is $3.93.

The total cost is $23.70.

the values are actually 19.7641, 3.9313, 23.6954 in the workspace.

How do I stop it from rounding the hidden numbers?

Answer 1

[Image Analyst]

Use more decimal places of precision if you want. Instead of 2 with $%0.2f you can use 6 with $%0.6f.

[Belle Dionido]

But I need it to be 2 decimal places because it is money value. My issue is that it's supposed to be $23.69, $19.76 + $ 3.93 = $23.69, not $23.70.

[Image Analyst]

What is the value of total?

What do you see when all 6 decimal places are printed out? The %.2f should be rounding correctly. Are you saying it rounds up when it shouldn't?

Maybe you want to try

value = round(value, 2) 

or use

value = 100 * floor(value*100)

[Belle Dionido]

Yes, that worked thank you!

[Image Analyst]

OK, great, but could you click the "Accept this answer" link? Thanks in advance. 🙂

[DGM]

Since it's not really clear which behavior you want, note that the behavior of floor() and fix() differ and may matter if you process negative inputs.

Answer 2

[DGM]

If you simply want to truncate the values to integer cents, consider the example:

A = [19.7641, 3.9313, 23.6954]

A = 1×3

19.7641 3.9313 23.6954

B = truncatecents(A)

B = 1×3

19.7600 3.9300 23.6900

fprintf('before truncation: %.2f\n',A(3))

before truncation: 23.70

fprintf('after truncation: %.2f\n',B(3))

after truncation: 23.69

function out = truncatecents(in)
    out = fix(in*100)/100; % truncate toward zero
end