Contradiction of variable existence

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?

[Dyuman Joshi]

I don't see a problem, 0! is 1 only.

Joke aside, using exist is the right way.

All syntaxes are working properly here -

y = 5;
exist('y')

ans = 1

exist y

ans = 1

exist('y', 'var')

ans = 1

if ~exist('x')
    x = 5;
    fprintf('x = %d', x);
else frpintf('x already exists');
end

x = 5

Is the variable deleted/cleared from the current workspace in between the exist calls?

Also note the difference between the synatxes - Command vs Function Syntax

Please share your code so that we can reproduce the issue.

[Valeriy]

Thank you Joshi, answer on my question are single or doble quotes between which MUST be name of the variable. Without them command "exist" is not correct.

BTW, attention to the MATLAB staff, see, please corresponding help for "exist". Example with quotes is in the end of the second help screen! This is the real reason of my question. It seems that such example must be placed in the first strings of the help article.

[Valeriy]

Sorry, I forgot where is "Accept this answer" button?

[Voss]

"answer on my question are single or doble quotes between which MUST be name of the variable... where is "Accept this answer" button?"

@Valeriy: You can Accept my answer below, since I said exactly that (the name of the variable must be supplied to the exist function).

[Valeriy]

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

[Voss]

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?

Answer 1

[Voss]

Use:

~exist("strF","var")

[Voss]

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

[Voss]

And the reason that this works

exist strF

is because it is command syntax. See:

https://www.mathworks.com/help/matlab/matlab_prog/command-vs-function-syntax.html