We usually use tf.cond op to express a conditional branching. However, its semantics are not that intuitive. Let’s see some toy examples:

def study_cond():
    a = tf.get_variable(name='a', shape=(), initializer=tf.constant_initializer(.0))
    op0 = tf.assign_add(a, 1.0)
    op1 = tf.assign_add(a, -1.0)
    p = tf.placeholder(dtype=tf.bool, shape=(), name='p')
    where_op = tf.where(p, op0, op1)
    b = tf.placeholder(shape=(), dtype=tf.float32)
    cond_op0 = tf.cond(p, lambda: op0, lambda: op1)
    cond_op1 = tf.cond(p, lambda: tf.assign_add(a, b), lambda: tf.assign_add(a, -1.0))

    with tf.Session() as sess:
        for i in range(1024):
            sess.run(where_op, feed_dict={p: False})
        print(a.eval(sess))# a is .0
        for i in range(1024):
            sess.run(cond_op0, feed_dict={p: False})
        print(a.eval(sess))# a is .0
        for i in range(1024):
            sess.run(cond_op1, feed_dict={p: False, b: 1.0})
        print(a.eval(sess))# a is -1024

As you can see, unless the assignment ops are defined inside the callables, they will be executed no matter in which branch they are called. If you are supriced, official document of TF explains this as the nature of Dataflow programming model and your expectation is a so-called lazier semantics. This thread gives a vivid explanation:

Because execution in a TensorFlow graph flows forward through the graph, all operations that you refer to in either branch must execute before the conditional is evaluated.

Note that in our example, if we run the cond_op1 op without feeding a value for placeholder b, the program will crash.