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

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Belle Dionido
Belle Dionido 2022-2-16
评论： DGM 2022-2-16
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?

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采纳的回答

Image Analyst
Image Analyst 2022-2-16
Use more decimal places of precision if you want. Instead of 2 with $%0.2f you can use 6 with $%0.6f.
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DGM
DGM 2022-2-16
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.

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更多回答（1 个）

DGM
DGM 2022-2-16
编辑：DGM 2022-2-16
Ran in:
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

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