clear all; close all; clc;
x = input('x= ');
y = input('y= ');
if (x>=5) & (y>2)
f(x,y)=x^2-y;
elseif (x<5) | (y>0 & y<=2)
f(x,y) = x-y^2;
else (x<0 & x>0) & (y<0)
f(x,y)= x^3 + y^3;
end
f(x,y)
%when I try input x =0 and y=2 it does not run
%Index in position 1 is invalid. Array indices must be positive integers or logical values.
%How to I solve that?
%Thank you very much

1 Kommentar

clear all; close all; clc;
x = 0 ;
y = 2 ;
if (x>=5) & (y>2)
f = @(x,y) x^2-y; % define function using function handle
f(x,y)
elseif (x<5) | (y>0 & y<=2)
f= @(x,y) x-y^2;
f(x,y)
else (x<0 & x>0) & (y<0)
f = @(x,y) x^3 + y^3;
f(x,y)
end
ans = -4
Define the function using its function handle as the way you try to call function

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Matt J
Matt J am 2 Jan. 2023
Bearbeitet: Matt J am 2 Jan. 2023

0 Stimmen

x=0; y=2;
if (x>=5) & (y>2)
f=x^2-y;
elseif (x<5) | (y>0 & y<=2)
f = x-y^2;
else (x<0 & x>0) & (y<0)
f= x^3 + y^3;
end
f
f = -4

2 Kommentare

Vo
Vo am 2 Jan. 2023
why we use f due to f(x,y) sir ?
Matt J
Matt J am 2 Jan. 2023
Bearbeitet: Matt J am 2 Jan. 2023
Because, "array indices must be positive integers or logical values"
For example:
A=rand(1,4)
A = 1×4
0.8551 0.1501 0.7576 0.0665
We can do
A(1)
ans = 0.8551
A(2)
ans = 0.1501
but not,
A(0)
Array indices must be positive integers or logical values.
There is no location in the vector known as A(0).

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Vo
am 2 Jan. 2023

Bearbeitet:

am 2 Jan. 2023

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