how I do it. the function is called a string is passes a numbr
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Dezdi
el 19 de Nov. de 2018
Comentada: dpb
el 20 de Nov. de 2018
how could I do this one
looks like
functionname('hello')
ans= hello1
functionname('john')
ans= john2
...
many thanks
1 comentario
dpb
el 19 de Nov. de 2018
Which number, in particular? Any one, random, user-specified, sequential since last invocation as shown above (requires some extra logic to keep track of that and over what population if so)...???
So many questions, so little known... :)
s=sprintf('%s%d',str,randi(1000);
might give you some ideas, however...
Respuesta aceptada
Stephen23
el 20 de Nov. de 2018
function out = myfun(inp)
persistent cnt
if nargin==0 || isempty(cnt)
cnt = 0;
end
cnt = cnt+1;
assert(cnt<=5,'You already have five words')
out = sprintf('%s%d',inp,cnt);
end
And tested:
>> myfun('anna')
ans = anna1
>> myfun('bob')
ans = bob2
>> myfun('cathy')
ans = cathy3
>> myfun('david')
ans = david4
>> myfun('evalyn')
ans = evalyn5
>> myfun('frog')
error: You already have five words
1 comentario
Más respuestas (2)
Luna
el 20 de Nov. de 2018
Hello Dezdi,
You can do it by first collecting your inputs in a cell array and then call your function in a for loop with 5 elements.
inputStr = {'John','Mary','Kayle','Adam','David'};
for i = 1:5
out = myFunc(inputStr,i)
end
function outputStr = myFunc(inputStr,i)
outputStr = [inputStr{i}, sprintf('%d',i)];
end
Preethi
el 20 de Nov. de 2018
hi,
check this
function test_preethi(str)
persistent i
if isempty(i)
i=1;
else
if i <=5
sprintf('%s%d',str,i)
i=i+1 ;
else
sprintf('limit reached')
end
end
3 comentarios
dpb
el 20 de Nov. de 2018
This particular function only "works" in that it creates the string internal to the function but you can't get at the result.
The use of persistent is good in saving the previous value, unfortunately the function has no mechanism by which the variable can be reset/re-initialized so without external intervention once the sequence is full, there's no going forward...details, details! :)
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