How to write to screen a set of variables with mixed types?

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Robert Jones
Robert Jones 2024-12-30
回答： Walter Roberson 2025-1-4
Ran in:
hello,
I am trying to write to screen a set of variables with mixed types. I have the code below
and I get the following display:
d=0.7874 mm , epsr=2.1 >>
a='d=0.7874 mm , epsr=2.1';
m=3;
k=7;
u=7.853535;
fprintf('%s %i \t %10.5f \n',[a m u])
d=0.7874 mm , epsr=2.1□□
help would be appreciated.
Thank you

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回答（2 个）

Manikanta Aditya
Manikanta Aditya 2024-12-30
编辑：Manikanta Aditya 2024-12-30
Ran in:
HI @Robert Jones
It looks like you're trying to print a string along with integer and floating-point variables using fprintf. The issue arises because you're trying to concatenate different types into a single array, which fprintf doesn't handle well. Instead, you should pass each variable separately to fprintf.
Here's how you can modify your code to correctly display the variables:
a = 'd=0.7874 mm , epsr=2.1';
m = 3;
k = 7;
u = 7.853535;
% Use fprintf with separate arguments for each variable
fprintf('%s %i \t %10.5f \n', a, m, u);
d=0.7874 mm , epsr=2.1 3 7.85353
This will correctly print the string a, the integer m, and the floating-point number u in the desired format.
If you want to include the variable k as well, you can add it to the fprintf statement:
fprintf('%s %i \t %i \t %10.5f \n', a, m, k, u);
d=0.7874 mm , epsr=2.1 3 7 7.85353
This way, each variable is passed as a separate argument to fprintf, ensuring they are printed correctly.
Refer to the following doc:
I hope this helps.
Walter Roberson
Walter Roberson 2025-1-4
Ran in:
a='d=0.7874 mm , epsr=2.1';
m=3;
u=7.853535;
temp = [a m u]
temp = 'd=0.7874 mm , epsr=2.1□□'
length(a)
ans = 22
length(temp)
ans = 24
temp(end-1:end), temp(end-1:end)+0
ans = '□□'
ans = 1×2
3 7
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You are using [] to concatenate a character vector and numeric items. As described in https://www.mathworks.com/help/matlab/matlab_prog/valid-combinations-of-unlike-classes.html the result of concatenating a character vector and numeric items is a character vector, with each of the numeric items having first been converted to uint16() after fractions are dropped. The numeric values are not first converted to printable representations: they are treated as code-points.
So you start with your vector of length 22, and concatenate on two code points, giving a result that is a vector of length 24, the last two code points of which are 3 and 7.
Your error was in concatenating the values first, instead of using individual parameters to fprintf()

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