Use eval function with strings containing if clauses and for loops

Hi,

this is what I'm trying to do:

1. Build a string containing code (including a for loop and an if clause)
2. Execute the string as if it were actual code using the eval() function

Example:

a=5;
eval('if a>4 b=1 else b=0 end')

The error displayed is:

Error: Illegal use of reserved keyword "else".

If I remove the else and execute the lines:

a=5;
eval('if a>4 b=1 end')

I get:

Error: Illegal use of reserved keyword "end".

The same error appears if the string contains a for loop.

Is there a way to bypass the problem? Maybe another function I am not aware of?

Thank you!

Answer 1

[Mina Mohaghegh]

this example works, I believe you can figure out the rest:

eval(char(sprintf('for i=1:2 i \n end ')))

Answer 2

[Michael Doroginizky]

>> eval('S=0')

S =

0

>> eval('S=0; for i = 1 : 10 S=S+1; end')

>> S

S =

10

>> eval('S=0; if S > 0 S=5; elseif S < -1 S=7; else S=10; end')

>> S

S =

10

[Rik]

Why would you encourage the use of eval by reviving such an old thread?

[Steven Lord]

If you're using eval so you can execute this code inside an anonymous function, for example, you don't need to. Taking a look at the most complicated of the three:

eval('S=0; if S > 0 S=5; elseif S < -1 S=7; else S=10; end')

That could be replaced with a call to discretize (which would have the added benefit of being able to handle a non-scalar S array.)

S = -5:5;
x = discretize(S, [-Inf -1 0 Inf], [7, 10, 5]);
[S; x]

ans = 2×11

-5 -4 -3 -2 -1 0 1 2 3 4 5 7 7 7 7 10 5 5 5 5 5 5

For values of S less than 1 (and greater than or equal to -Inf), the corresponding elements of x are 7. For values greater than or equal to 0 (and less than or equal to Inf) the corresponding elements of x are 5. Otherwise (in this case S = -1) the corresponding elements of x are 10.

If you really need that clause to be strictly greater than, change the 0 in the discretize call to something small but nonzero like eps or eps(0).

x2 = discretize(S, [-Inf -1 eps Inf], [7, 10, 5]);
x3 = discretize(S, [-Inf -1 eps(0) Inf], [7, 10, 5]);
[S; x2; x3]

ans = 3×11

-5 -4 -3 -2 -1 0 1 2 3 4 5 7 7 7 7 10 10 5 5 5 5 5 7 7 7 7 10 10 5 5 5 5 5