stack

来源： https://stackoverflow.com/questions/47112183/rsp-doesnt-move-when-entering-new-function

来源： https://stackoverflow.com/questions/37749269/is-the-stack-in-a-cache

来源： https://stackoverflow.com/questions/31673065/creating-classes-in-c-on-the-stack-vs-the-heap

来源： https://stackoverflow.com/questions/31673065/creating-classes-in-c-on-the-stack-vs-the-heap

问题 It's a little bit confusing. Is it actually comes from stack pop/push terminology? L.pop([index]) -> item -- remove and return item at index (default last). Raises IndexError if list is empty or index is out of range. >>> [1,2].pop() 2 Remove and return an arbitrary set element. Raises KeyError if the set is empty. >>> {1,2}.pop() 1 回答1: set.pop does not return the first element. As it clearly says in the help that you quoted, it returns an arbitrary element. And the reason for this is simple

问题 It's a little bit confusing. Is it actually comes from stack pop/push terminology? L.pop([index]) -> item -- remove and return item at index (default last). Raises IndexError if list is empty or index is out of range. >>> [1,2].pop() 2 Remove and return an arbitrary set element. Raises KeyError if the set is empty. >>> {1,2}.pop() 1 回答1: set.pop does not return the first element. As it clearly says in the help that you quoted, it returns an arbitrary element. And the reason for this is simple

问题 It's a little bit confusing. Is it actually comes from stack pop/push terminology? L.pop([index]) -> item -- remove and return item at index (default last). Raises IndexError if list is empty or index is out of range. >>> [1,2].pop() 2 Remove and return an arbitrary set element. Raises KeyError if the set is empty. >>> {1,2}.pop() 1 回答1: set.pop does not return the first element. As it clearly says in the help that you quoted, it returns an arbitrary element. And the reason for this is simple

问题 NASM Assembly, Ubuntu, 32-bit program. Normally, when popping a value from the stack, I'll do POP somewhere Into a register or a variable. But sometimes, I simply don't want to put it anywhere - I just want to get rid of the next element in the stack. Doing POP Just like that won't work. A workaround I had was to make a 4-byte variable I don't use at all and dump the POP into it. Is there a better way to achieve this? 回答1: Adjust the stack pointer by four bytes (or some other amount),

问题 This question already has answers here : Why does std::getline() skip input after a formatted extraction? (3 answers) Closed 2 months ago . Given an expression string exp. Examine whether the pairs and the orders of “{“,”}”,”(“,”)”,”[“,”]” are correct in exp. For example, the program should print 'balanced' for exp = “[()]{}{[()()]()}” and 'not balanced' for exp = “[(])” Input The first line of input contains an integer T denoting the number of test cases. Each test case consists of a string

问题 I have two arrays: A = np.array([1, 2, 3]) B = np.array([2, 3, 4]) C = np.stack((A, B), axis=0) print C.shape (2, 3) Shouldn't the shape be (6,) ? 回答1: Using the np.stack() function you can specify which axis would you like to be considered the index axis. So as you can see you will never get a shape of 6 , only (2,3) or (3,2) for this example depending on what axis you chose. See below: A = np.array([1, 2, 3]) B = np.array([2, 3, 4]) arrays = [A, B] With this code: print(np.stack(arrays,