Question

我收到两条错误消息：

Error: format '%d' expects a matching 'int' argument -wformat

Error: format '%f' expects an arguemnt of double but argument 4 has type int -wformat

我抬起头来试图解决它，但没有用。以下是我的代码有人能告诉我我做错了吗？

  #include <stdio.h>       
  int main() {
    int n = 5, a[5] = {1, 3, 5, 7, 9};  // Declare & initialize array of length 5
    int sum;
    int i;
    for (i = 1; i < n; i++) {
        sum = sum + a[i];

    printf("Enter an integer x: %d");       // Prompt the user
    int x;
    scanf("%d", &x);        // Read in the integer

    // Print out the sum, the division we're performing, and the result (without truncation)
    // E.g., The sum of the array is 25; 25/2 = 12.500000

    printf("The sum of the array is $d; %d/%d = %f\n", sum, sum, sum / x);

    // Declare an integer variable y and initialize it using a hexadecimal constant.
    // Print y in decimal, hex, and with leading zeros so that we get the output
    // y = 4011 = fab = 0xfab =   fab = 0000fab

    int y = 0xfab;
    printf("y = %d = %x\n", y, y, y, y, y);
    return 0;

Answer 1

您正在以错误的方式使用printf：

 printf("Enter an integer x: %d");

您必须指定要在％d出现时打印的整数值，如下所示：

 printf("Enter an integer x: %d",someValue);

这也错了：

 printf("The sum of the array is $d; %d/%d = %f\n", sum, sum, sum / x);

您正在使用％f打印整数。您应该执行以下操作：

 printf("The sum of the array is $d; %d/%d = %f\n", sum, sum, ((double)sum / (double)x));

Answer 2

这是错误的：

printf("The sum of the array is $d; %d/%d = %f\n", sum, sum, sum / x);

我相信你想要这个：

printf("The sum of the array is $d; %d/%d = %d\n", sum, sum, sum/x);

请注意，上一个%f变为%d，因为sum/x将产生一个整数。

这也是错误的：

printf("Enter an integer x: %d");

删除%d，你不能传递格式说明符，然后错过相应的参数。

错误：格式＆＃39;％d＆＃39;期待匹配的＆＃39; int＆＃39;参数-wformat

2 个答案: