小编use*_*456的帖子

我正试图argv在我的代码中打印出此时的值,但是当我尝试编译它时,我得到了 warning: format ‘%p’ expects argument of type ‘void *’, but argument 3 has type ‘char **.我不确定这个警告意味着什么.是%p不是要用于argv即使它是一个指针？

int main(int argc, char* argv[])
{
    printf("%s%p", "argv = ", argv);
}

即使"退出"或num输入,程序也不会打破while循环.在函数的底部,我尝试使用guess == num,并返回true.最底层函数外部的代码是该函数的基本结构,并且该函数正常工作.不打破while循环的问题是否与函数中的两个for循环有关？谢谢.

def cowsAndBulls():  
    guess = ''  
    num = str(random.randint(1000, 9999))

    while guess != num or guess != 'exit':    
        guess = input('Please enter a guess for a 4-digit number (\'exit\' to leave): ')
        numL = [digit for digit in num]
        guessL = [digit for digit in guess]
        cow = 0
        bull = 0

        for digit in guessL:
            if digit in numL and guessL.index(digit) == numL.index(digit):
                cow = cow + 1
                guessL[guessL.index(digit)] = 'G'
                numL[numL.index(digit)] = …