Placeholder_2:0 is both fed and fetched

Question

When I run this code:

x = tf.placeholder(tf.int32, shape=(None, 3))
with tf.Session() as sess: 
    feed_dict = dict()
    feed_dict[x] = np.array([[1,2,3],[4,5,6],[7,8,9],[10,11,12]])
    input = sess.run([x], feed_dict=feed_dict)

I get this error:

Placeholder_2:0 is both fed and fetched.

I'm not sure what I'm doing wrong here. Why does this not work?

Possible duplicate of [TensorFlow - Invalid argument: Reshape:0 is both fed and fetched](https://stackoverflow.com/questions/37076208/tensorflow-invalid-argument-reshape0-is-both-fed-and-fetched) — buzjwa, Sep 17 '18 at 10:44

score 10 · Accepted Answer · answered Sep 03 '16 at 13:12

10

Are you sure this code covers what you are trying to achieve? You ask to read out whatever you pass through. This is not a valid call in tensorflow. If you want to pass through values and do nothing with it (what for?) you should have an identity operation.

x = tf.placeholder(tf.int32, shape=(None, 3))
y = tf.identity(x)

with tf.Session() as sess: 
    feed_dict = dict()
    feed_dict[x] = np.array([[1,2,3],[4,5,6],[7,8,9],[10,11,12]])
    input = sess.run([y], feed_dict=feed_dict)

The problem is "feeding" actually kind of overwrites whatever your op generates, thus you cannot fetch it at this moment (since there is nothing being really produced by this particular op anymore). If you add this identity op, you correctly feed (override x) do nothing with the result (identity) and fetch it (what identity produces, which is whatever you feeded as an output of x)

answered Sep 03 '16 at 13:12

lejlot

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Yes. You are correct! I realized what I was doing wrong and posted an answer as you were also posting this answer. Thank you for taking the time to post your answer. I understand now – Taivanbat Badamdorj Sep 03 '16 at 13:15
I was trying to run `x` inside of `sess.run()` when there was nothing to run – Taivanbat Badamdorj Sep 03 '16 at 13:15
As a side question, if we use `y = x` instead of `y = tf.identity(x)` will making changes to `y` change `x`? – Taivanbat Badamdorj Sep 03 '16 at 13:17
1

y = x does nothing, this is simply a pure python code which assigns reference of x to variable y, this does nothing to underlying graph. Thus the answer is no, this will not work. And yes any change to y will affect x as well (since this is a reference to the same object). – lejlot Sep 03 '16 at 13:19
Ah. That makes sense – Taivanbat Badamdorj Sep 03 '16 at 13:26

score 2 · Answer 2 · answered Sep 03 '16 at 13:13

I found out what I was doing wrong.

x is a placeholder -- it holds information and evaluating x does not do anything. I forgot that vital piece of information and proceeded to attempt to run the Tensor x inside sess.run()

Code similar to this would work if, say, there was another Tensor y that depended on x and I ran that like sess.run([y], feed_dict=feed_dict)

Placeholder_2:0 is both fed and fetched

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