C 程序只对第一个提示进行输入

Question

My program only takes input for the first prompt and it excludes the succeeding ones, and instead, it just directly prints them.

#include <stdio.h>

int main(){
    
    // Initialize
    char cCurrencyA, cCurrencyB;
    float fCurrencyA;
    float fRate;
    
    // Prompt user for currency names
    printf("Enter a currency name: ");
    scanf("%c%c", &cCurrencyA, &cCurrencyB);
    
    // Prompt user for the current rate
    printf("Enter the rate:");
    scanf("%f", &fRate);
    
    // Prompt user to enter value for currency A
    printf("Enter value for %c", cCurrencyA);
    scanf("%f", &fCurrencyA);
    
    // Convert currency A to B
    int nResult = fCurrencyA * fRate;
    
    // Print the result
    printf("%f%c is %d%c", fCurrencyA, cCurrencyA, nResult, cCurrencyB);
    return 0;
    
}

Output:

Enter a currency name:USD PHP
Enter the rate:
Enter value for U:
0.000000U is 0S

Answer 1

1. so you have major problems, you are scanning a whole string into only one char , that's wrong, you should do char cCurrencyA[20], cCurrencyB[20]; not char cCurrencyA, cCurrencyB; to scan a string not only one char .

2. when you are scanning using scanf() , use the quantifier %s not %c , as %c will get only one character but %s will scan a whole string.

3. also to round the result, you could use math header file, by using the function called lround() , so do this int nResult = round((double)fCurrencyA * fRate); not int nResult = fCurrencyA * fRate; as the lValue is int while RValue is float so you have to cast it.

and here the code edited:

#include <stdio.h>
#include <math.h>
int main(){

    // Initialize
    char cCurrencyA[20], cCurrencyB[20];
    float fCurrencyA;
    float fRate;

    // Prompt user for currency names
    printf("Enter a currency name: ");
    scanf("%19s %19s", cCurrencyA, cCurrencyB);

    // Prompt user for the current rate
    printf("Enter the rate:");
    scanf("%f", &fRate);

    // Prompt user to enter value for currency A
    printf("Enter value for %s", cCurrencyA);
    scanf("%f", &fCurrencyA);

    // Convert currency A to B
    int nResult = lround((double)fCurrencyA * fRate);

    // Print the result
    printf("%f%s is %d%s", fCurrencyA, cCurrencyA, nResult, cCurrencyB);
    return 0;

}

and here is the output:

Enter a currency name:USD PHP
Enter the rate:10.5
Enter value for USD10
10.000000USD is 105PHP
Process finished with exit code 0

Answer 2

First line of input was 8 characters: "USD PHP\n" .

scanf("%c%c", &cCurrencyA, &cCurrencyB); only reads and saves the first 2 in cCurrencyA , cCurrencyB .

The remaining 6 are not numeric text for the next scanf("%f",... so it does not assign anything and returns 0. Same for the next scanf("%f",... .

---

Best to read a line of user input with fgets() and then parse the string.

---

Yet if one must use the problematic scanf() , read currency names as a string.

// char cCurrencyA, cCurrencyB;
char cCurrencyA[100], cCurrencyB[100];

// scanf("%c%c", &cCurrencyA, &cCurrencyB);
scanf("%99s %99s", cCurrencyA, cCurrencyB);

// printf("%f%c is %d%c", fCurrencyA, cCurrencyA, nResult, cCurrencyB);
printf("%f%s is %d%s\n", fCurrencyA, cCurrencyA, nResult, cCurrencyB);

Answer 3

When using scanf with the %c conversion format specifier, scanf will only match a single character. Therefore, with the input USD PHP , the function call

scanf("%c%c", &cCurrencyA, &cCurrencyB);

will match the U and the S , but D PHP will be left on the input stream. Afterwards, the function calls

scanf("%f", &fRate);

and

scanf("%f", &fCurrencyA);

will both fail, because scanf is unable to convert D PHP to a floating-point number.

If you want to use scanf to match a whole word, you could use the %s format specifier instead, but you should normally limit it to the size of the memory buffer. For example, if the size of the memory buffer is 4 , then you could use %3s as the format specifier, which limits the number of matched characters to 3 , so that you still have room for the terminating null character.

However, it is generally not recommended to use scanf for line-based user input, because, as you have now discovered yourself, scanf does not always behave in an intuitive manner, because it does not always read one line at once.

For this reason, you may want to consider using the fgets function instead, which always reads exactly one line of input at once, if possible.

Therefore, I recommend that you first read an entire line of input as a string using fgets , and then use strtof , sscanf and/or maybe strtok to parse the string:

#include <stdio.h>
#include <stdlib.h>

int main( void )
{
    char line[100];
    char strCurrencyA[4], strCurrencyB[4];
    float fCurrencyA, fRate, fResult;
    
    // Prompt user for currency names
    printf( "Enter a currency name: " );
    fgets( line, sizeof line, stdin );
    sscanf( line, "%3s %3s", strCurrencyA, strCurrencyB );
    
    // Prompt user for the current rate
    printf( "Enter the rate: " );
    fgets( line, sizeof line, stdin );
    fRate = strtof( line, NULL );
    
    // Prompt user to enter value for currency A
    printf("Enter value for %s: ", strCurrencyA);
    fgets( line, sizeof line, stdin );
    fCurrencyA = strtof( line, NULL );
    
    // Convert currency A to B
    fResult = fCurrencyA * fRate;
    
    // Print the result
    printf( "%0.2f %s is %0.2f %s", fCurrencyA, strCurrencyA, fResult, strCurrencyB);

    return 0;    
}

This program has the following behavior:

Enter a currency name: USD PHP
Enter the rate: 56.8
Enter value for USD: 25
25.00 USD is 1420.00 PHP

Note that this program does not perform any input validation. It will probably misbehave if the input is invalid in any way.