Comparison of numbers read with "io.read()" does not result in true as expected

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-2

I tried to put 1 and the second print was not. If I change x=io.read() for "x=1" it works normally

Follows the code

x = io.read()
if x ~= 0 then
    print(x.." diferente de 0")
end
if x == 1 then
    print(x.." igual a 1")
end

Example of input and output:

1
1 different than 0

if typing lua
  • 1

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  • 1

    in lua when you give an io.read the input comes a string...so comparing it is "1" == 1, so it is false, to work out you could for example convert this io.read to integer with tonumber, ex : x = tonumber(io.read())

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2 answers

4

Must be:

x = io.read()
if x ~= '0' then
    print (x .. ' deferente do 0')
end
if x == '1' then
    print (x .. ' igual a 1')
end

io.read() returns strings.

4

You are comparing text with number, so they are different values, so it is false and does not enter the if. That is correct:

x = io.read()
if x ~= 0 then
    print(x .. " diferente de 0")
end
if x == "1" then
    print(x .. " igual 1")
end
if tonumber(x) == 1 then
    print(x .. " igual 1")
end

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

Either you compare text with text, or convert text to number to compare with number.

I didn’t get into the merit that the conversion could go wrong, and so it would need to be tested before using, so it might not be a good one used simply like this.

The first if gives true because "1" is different from 0, they are of different types, so do not even need to look at the value of each. This is the correct comparison.

Can read more about the function used that always returns string.

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