c++ - 将数组作为参数传递给 printf

Question

有没有一种方法可以执行以下操作，但只将 bounds 传递给 printf？

double *bounds = getBounds();
printf("%f-%f, %f-%f, %f-%f",
    bounds[0], bounds[1],
    bounds[2], bounds[3],
    bounds[4], bounds[5] );

// what I'd like to write instead:
xxxprintf("%f-%f, %f-%f, %f-%f", bounds);

Answer 1

你可以自己写xxxprintf()

#include <stdio.h>

int arrayprintf_dbl(const char *fmt, const double *data) {
  int n = 0;
  const char *p = fmt;
  const double *x = data;
  while (*p) {
    if (*p == '%') {
      // complicate as needed ...
      p++;
      if (*p != 'f') return -1; // error
      n += printf("%f", *x++);
    } else {
      putchar(*p);
      n++;
    }
    p++;
  }
  return n;
}

int main(void) {
  double bonus[6] = {1, 2, 3, 4, 5, 6};
  arrayprintf_dbl("%f-%f, %f-%f, %f-%f\n", bonus);
  return 0;
}

^{我用 C 写的，我认为它可以很容易地转换为 C++(我不懂 C++)。}