String Character mixup on data types
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Hey all, so I used sprintf to create a char. When I look at the workspace at it's class it shows as character however the value is 1x48 char. When I copy the output of that same item and set test= 'text' I see a class of character however the value is 'text' or 'text... (for longer strings).
Why are these not the same? What extremely simple thing am I forgetting in my tired state?
Thanks!
Can provide more input if needed!
7 件のコメント
Geoff Hayes
2015 年 3 月 8 日
Karl - please show the code that you are using to do create the 1x48 array of characters, and the code that you are using to initialized test. If you are setting the latter as
test = 'text'
then it makes sense that the variable test takes on the string/character value of 'test' because you are assigning the latter (a string and not a variable) to the former. Why would you expect this to be the same as the 1x48 char array? Is text a variable? If so then you don't need to wrap it in single quotes in order to assign it to text.
Geoff Hayes
2015 年 3 月 8 日
Karl - infoline is a string as well (due to the sprintf). You may want to remove the \n and \b (?) from your sprintf command to get something like
infoline = sprintf('%s %s%dnm %dgra %ds(%dx) %dFrame',header.date,header.ExperimentTimeLocal,header.SpecCenterWlNm,header.SpecGrooves,header.exp_sec,header.lavgexp,header.NumFrames);
and try this with your call to title.
Karl
2015 年 3 月 8 日
Geoff Hayes
2015 年 3 月 8 日
Or maybe the problem lies with the header.date and header.ExperimentTimeLocal. Have you verified that both are of data type char (strings)?
Karl
2015 年 3 月 9 日
Geoff Hayes
2015 年 3 月 10 日
Karl - ischar is sufficient as is using class. What happens if you try the following to narrow down the problem
infoline1 = sprintf('%s',header.date);
infoline2 = sprintf('%s %s ',header.date,header.ExperimentTimeLocal);
infoline3 = infoline = sprintf('%s %s%dnm ',header.date,header.ExperimentTimeLocal,header.SpecCenterWlNm);
Do all produce reasonable strings? I can't help but wonder if there is a problem with one of your inputs (not a string, not an integer, etc.).
回答 (1 件)
Jan
2015 年 3 月 8 日
0 投票
The code is correct in both cases. It is only the display in the command window which appears under certain conditions in the abbreviated form. The contents of the variable is a char string in both cases, so everything is fine.
If you post the code, which reproduces the problem, it would be easy to demonstrate what's going on.
1 件のコメント
Guillaume
2015 年 3 月 8 日
And this is not particular to strings. For short vectors of anything, matlab displays the content of the vectors in the workspace browser. For longer vectors, it just displays their size.
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