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Exponentiation with infix operators in (or operating on) the base: Difference between revisions
Exponentiation with infix operators in (or operating on) the base (view source)
Revision as of 17:57, 4 November 2020
, 3 years agoAdd Python
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───── ──────╨─────────── ───────╨─────────── ───────╨─────────── ───────╨─────────── ──────
</pre>
=={{header|Python}}==
<lang python>from itertools import product
xx = '-5 +5'.split()
pp = '2 3'.split()
texts = '-x**p -(x)**p (-x)**p -(x**p)'.split()
print('Integer variable exponentiation')
for x, p in product(xx, pp):
print(f' x,p = {x:2},{p}; ', end=' ')
x, p = int(x), int(p)
print('; '.join(f"{t} =={eval(t):4}" for t in texts))
print('\nBonus integer literal exponentiation')
X, P = 'xp'
xx.insert(0, ' 5')
texts.insert(0, 'x**p')
for x, p in product(xx, pp):
texts2 = [t.replace(X, x).replace(P, p) for t in texts]
print(' ', '; '.join(f"{t2} =={eval(t2):4}" for t2 in texts2))</lang>
{{out}}
<pre>Integer variable exponentiation
x,p = -5,2; -x**p == -25; -(x)**p == -25; (-x)**p == 25; -(x**p) == -25
x,p = -5,3; -x**p == 125; -(x)**p == 125; (-x)**p == 125; -(x**p) == 125
x,p = +5,2; -x**p == -25; -(x)**p == -25; (-x)**p == 25; -(x**p) == -25
x,p = +5,3; -x**p ==-125; -(x)**p ==-125; (-x)**p ==-125; -(x**p) ==-125
Bonus integer literal exponentiation
5**2 == 25; - 5**2 == -25; -( 5)**2 == -25; (- 5)**2 == 25; -( 5**2) == -25
5**3 == 125; - 5**3 ==-125; -( 5)**3 ==-125; (- 5)**3 ==-125; -( 5**3) ==-125
-5**2 == -25; --5**2 == 25; -(-5)**2 == -25; (--5)**2 == 25; -(-5**2) == 25
-5**3 ==-125; --5**3 == 125; -(-5)**3 == 125; (--5)**3 == 125; -(-5**3) == 125
+5**2 == 25; -+5**2 == -25; -(+5)**2 == -25; (-+5)**2 == 25; -(+5**2) == -25
+5**3 == 125; -+5**3 ==-125; -(+5)**3 ==-125; (-+5)**3 ==-125; -(+5**3) ==-125</pre>
=={{header|Wren}}==
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