Question

很抱歉，如果这是一个愚蠢的问题，我是C ++的新手。为什么不能正确地将所有输入复制到输出中？

#include <iostream>
#include <iomanip>
#include <limits>
#include <string>

using namespace std;

int main() {
    int num1;
    long num2;
    long long num3;
    char char1;
    float num4;
    double num5;
    scanf("%d %ld %lld %c %f %lf", &num1, &num2, &num3, &char1, &num4, &num5);
    //input: 211916801 452082285 97592151379235457 p 19856.992 -5279235.721231465
    printf("%d %ld %lld %c %f %lf", num1, num2, &num3, &char1, &num4, &num5);
    //expected output: 211916801 452082285 97592151379235457 p 19856.992 -5279235.721231465
    //actual output: 211916801 452082285 68674564278975280 c 0.000000 0.000000
    system("pause");
    return 0;
}

Answer 1

因为某些变量的&不符合您的预期。

printf("%d %ld %lld %c %f %lf", num1, num2, &num3, &char1, &num4, &num5);

应该是

printf("%d %ld %lld %c %f %lf", num1, num2, num3, char1, num4, num5);

您的编译器应该已经警告过，因为您传递的值与格式说明符不匹配。如果没有，请打开/增加编译器警告。

Answer 2

您应该将值本身而不是指针传递给printf。

printf("%d %ld %lld %c %f %lf", num1, num2, num3, char1, num4, num5);

Answer 3

在Printf（大多数情况下）你不使用＆amp;在你的变量中。

printf("%d %ld %lld %c %f %lf", num1, num2, num3, char1, num4, num5);

另外，如果您是C ++的新手，可以采用更好的方式执行I / O操作。使用“cin”代替“scanf”和“cout”而不是“printf”

#include <iostream>
#include <iomanip>
#include <limits>
#include <string>

using namespace std;

int main() {
    int num1;
    long num2;
    long long num3;
    char char1;
    float num4;
    double num5;

   //scanf("%d %ld %lld %c %f %lf", &num1, &num2, &num3, &char1, &num4, &num5);
   cin >> num1;
   cin >> num2;
   cin >> num3;
   cin >> char1;
   cin >> num4;
   cin >> num5;
   //input: 211916801 452082285 97592151379235457 p 19856.992 -5279235.721231465

   //printf("%d %ld %lld %c %f %lf", num1, num2, &num3, &char1, &num4, &num5);
   cout << num1;
   cout << num2;
   cout << num3;
   cout << char1;
   cout << num4;
   cout << num5;

   //expected output: 211916801 452082285 97592151379235457 p 19856.992 -5279235.721231465
   //actual output: 211916801 452082285 68674564278975280 c 0.000000 0.000000

  system("pause");
  return 0;
}

Answer 4

使用printf(...)时，原始类型按值传递，而不是通过引用传递。使用以下代码：

#include <iostream>
#include <iomanip>
#include <limits>
#include <string>

using namespace std;

int main() {
    int num1;
    long num2;
    long long num3;
    char char1;
    float num4;
    double num5;
    scanf("%d %ld %lld %c %f %lf", &num1, &num2, &num3, &char1, &num4, &num5);
    printf("%d %ld %lld %c %f %lf", num1, num2, num3, char1, num4, num5);
    system("pause");
    return 0;
}

printf打印错误的char和0

4 个答案: