Printf打印错误的字符和0

Question

Sorry if this is a stupid question, I'm new to C++. Why is it not copying all of the input to the output correctly?

#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

Because you have & for some of the variables which is not what you intended.

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

should be

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

Your compiler should have warned about this as the values you pass don't match the format specifiers. If not, turn on/increase the compiler warnings.

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

In Printf(in most cases) you don't use & in your variables.

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

Plus, if you're new to C++, there's a better way to perform I/O Operations. Use "cin" instead of "scanf" and "cout" instead of "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

While using printf(...) , primitive types are passed by value not by reference. Use below code:

#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;
}