Question

我是C的新手，想要打印数组的一些特殊元素。

//Partial code here
    char aaa[18] ;
    int i = 0;
    while(i < 18) {
        aaa[i] = '#' ;
        printf("%c", aaa[i]) ;
        i = i + 4 ;    
   }

它应该打印4 # aaa [0]，aaa [4]，aaa [8]，aaa [12]，aaa [16]。但事实并非如此。它将它们打印成#####行。但我不想要那些。

Answer 1

我假设你想最终打印一个字符串并获得这样的输出

"#   #   #   #   #  "

减去报价。

你可以通过填写以空字符结尾的字符串然后printf来完成此操作，如下所示：

// +1 for space for a null terminator
// = {0}; fills the array with 0s
char aaa[18+1] = {0};

// for loop is more idiomatic for looping over an array of known size
for (int i = 0; i < 18; i += 4) {
    // if the remainder of dividing i by 4 is equal to 0
    if (i % 4 == 0) {
        // then put a '#' character in the array at aaa[i]
        aaa[i] = '#';
    } else {
        // otherwise put a ' ' character in the array at aaa[i]
        aaa[i] = ' ';
    }
}

printf("%s", aaa);

Answer 2

当您在每个循环中为i变量添加4时，您将连续打印阵列的位置0,4,8,12,16。

如果你想用＃作为4 * nth元素打印矢量，你应该做类似的事情：

while( i < 18 ) 
{
  if( i % 4 == 0 )
  {
    aaa[i] = '#';
    printf("%c" ,aaa[i]);
  }
  else 
    printf(" "); // Assuming you want a space in between prints

  i++;
}

部分在C

2 个答案: