Contradiction of variable existence

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Valeriy 2024-2-15

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https://ww2.mathworks.cn/matlabcentral/answers/2082598-contradiction-of-variable-existence

编辑： Voss 2024-2-15

Hello,

I met following situation. In workspace exist some string variable strF. I need to verify its existence and if it is not present, I should perform some actions. But I have following contradiction:

~exist(strF,"var") gives me logical 1, which means that strF don't exist. This is not correct.

But if I run command

exist strF

I have 1, that means that this variable exist. But when I try to realize this infomation by something like

TT = exist(strF) gives me 0!

How correctly determine existence of this variable?

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Valeriy 2024-2-15

@Voss name of the variable MUST BE suppied in quotes! Otherwise command's result is incorrect

Voss 2024-2-15

编辑：Voss 2024-2-15

Yes, the name of the variable, not the value of the variable. Name implies it's in quotes, when using function syntax! Did you follow the example I ran through in this comment?

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采纳的回答

Voss 2024-2-15

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https://ww2.mathworks.cn/matlabcentral/answers/2082598-contradiction-of-variable-existence#answer_1409628

在 MATLAB Online 中打开

Use:

~exist("strF","var")

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Voss 2024-2-15

在 MATLAB Online 中打开

You must supply exist with the name of the variable to check for.

This is what you were doing:

strF = "some_string";
% this checks for a variable called "some_string", which is the value of
% strF, but no variable called "some_string" exists
exist(strF,"var") 
ans = 0
some_string = 99; % now "some_string" exists, so
exist(strF,"var") % this returns true (1)
ans = 1

As opposed to:

clear("strF")       % clear strF
exist("strF","var") % strF doesn't exist now
ans = 0
strF = "some_string"; % now it does
exist("strF","var")   % and exist shows that
ans = 1
some_string = 99;   % and it doesn't matter about "some_string"
exist("strF","var") % because now exist is checking for a variable called "strF"
ans = 1