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    <p>It is not a problem of <code>do.call</code>, but <code>substitute</code> <s>which evaluate by default in the global environment. So</s> you need to tell it in which environment substitution must occur. Here obviously in the local envir of func3.</p> <p>This should work:</p> <pre><code> do.call("curve",list(expr = substitute(my.func, env = parent.frame()))) </code></pre> <p><strong>Edit</strong> thanks Dwin</p> <p>As said in the comment substitute env Defaults to the current evaluation environment. So Why the code below works? The answer in the help of <code>substitute</code></p> <blockquote> <p>formal argument to a function or explicitly created using delayedAssign(), the expression slot of the promise replaces the symbol. If it is an ordinary variable, its value is substituted, unless env is .GlobalEnv in which case the symbol is left unchanged.</p> </blockquote> <p><code>env = parent.frame(n=1)</code> is equivalent to <code>.GlobalEnv</code>, that why the symbol (my.func) is left unchanged. So the correct answer would be :</p> <pre><code>do.call("curve",list(expr = substitute(my.func, env = .GlobalEnv))) </code></pre> <p>To test , I open new R session :</p> <pre><code>func1 &lt;- function (m, n) { charac &lt;- paste ("func2 &lt;- function(x)", m, "*x^", n, sep = "") eval(parse(text = charac)) return(func2) } func3 &lt;- function (m, n) { my.func &lt;- func1 (m, n) do.call("curve",list(expr = substitute(my.func,env = .GlobalEnv))) } </code></pre> <p>Than I call </p> <pre><code> func3(2,6) </code></pre>
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    1. POdo.call and curve can not plot a function inside another function environment
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    1. USMatthew Plourde
    1. USagstudy
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    1. POdo.call and curve can not plot a function inside another function environment
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    1. COyes It works, thank you
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      1. USEhsan Masoudi
    2. CO@EhsanMasoudi if it works and you are satisfied of the question you can accept it by checking the box at left.
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    3. COThe help page for `substitute` says it is evaluated in the current environment.
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