how I do it. the function is called a string is passes a numbr
显示 更早的评论
how could I do this one
looks like
functionname('hello')
ans= hello1
functionname('john')
ans= john2
...
many thanks
1 个评论
dpb
2018-11-19
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...
采纳的回答
Stephen23
2018-11-20
0 个投票
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
更多回答（2 个）
Luna
2018-11-20
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
2018-11-20
0 个投票
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 个评论
Stephen23
2018-11-20
The question requests "...and returns the resulting string". Your code does not return anything.
dpb
2018-11-20
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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