Question

我正在为一个类做代码，一切看起来都应该有效，但是代码似乎不接受char输入和数学混乱。例如，如果我输入第一个输入1,000并输入'a'作为输入，它会重复询问输入，如果我将其自动设置为'a'，则会导致数学混乱。不知怎的，数学已经过时了。我该如何解决？任何帮助，将不胜感激。

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

void print_status_menu(void)
{
printf("MONTHLY        1       2       3       4       5       6       7       8\n");
printf("Status\n");
printf("A. Table for employees without qualified dependent\n");
printf("1. Z           1       0      833    2,500   5,833  11,667  20,833   41,667\n");
printf("2. S/ME        1     4,167   5,000   6,667  10,000  15,833  25,000   45,833\n\n");

printf("B. Table for  single/married employee with qualified dependent child(ren)\n");
printf("1. ME1 / S1    1     6,250   7,083   8,750  12,083  17,917  27,083   47,917\n");
printf("2. ME2 / S2    1     8,333   9,167   10,833 14,167  20,000  29,167   50,000\n");
printf("3. ME3 / S3    1    10,417   11,250  12,917 16,250  22,083  31,250   52,083\n");
printf("4. ME4 / S4    1    12,500   13,333  15,000 18,333  24,167  33,333   54,167\n");
}



void entered_filing_status(char *px, int *py, int*pz, float *pinc, int *ploc)
{
int def;
while(*px!='a' || *px!='b')
{
    printf("Table a or b?: ");
    *px = getchar();
    if(*px == 'A')
        *px = 'a';
    else if(*px == 'B')
        *px = 'b';
}

if(*px == 'b')

{
    while(*py<1)
    {
    printf("Input amount of children: ");
    scanf("%d", &*py);
    }
    while(*py>4)
    {
        *py-=1;
    }
    if(*pinc>(45833+(*py*2083)))
        def = 8;
    else if(*pinc>25000+(*py*2083))
        def = 7;
    else if(*pinc>15833+(*py*2083))
        def = 6;
    else if(*pinc>10000+(*py*2083))
        def = 5;
    else if(*pinc>6667+(*py*2083))
        def = 4;
    else if(*pinc>5000+(*py*2083))
        def = 3;
    else if(*pinc>4167+(*py*2083))
        def = 2;
    else
        def = 1;
}


else{
printf("Are you under zero exemptions?(0 for yes, 1 for no): ");
scanf("%d", &*pz);
*pz+=1;
if((*pinc)>(33333+(4167 * *pz)))
        def = 8;
    else if((*pinc)>12499+(4167 * *pz))
        def = 7;
    else if((*pinc)>3333+(4167 * *pz))
        def = 6;
    else if((*pinc)>(4167 * *pz)-2501)
        def = 5;
    else if((*pinc)>(4167 * *pz)-5834)
        def = 4;
    else if((*pinc)>(4167 * *pz)-7501)
        def = 3;
    else if((*pinc)>(4167 * *pz)-8334)
        def = 2;
}

*ploc = def;
}



void computed_tax(float *px, float *py, int *pz)
{
switch (*pz)
{
case 8:
    *py =  (10416 + (*px * 0.32));
    break;
    case 7:
    *py = (4166.67 + (*px * 0.3));
    break;
    case 6:
    *py = (1875 + (*px * 0.25));
    break;
    case 5:
    *py = (708.33 + (*px * 0.2));
    break;
    case 4:
    *py = (208.33 + (*px * 0.15));
    break;
    case 3:
    *py = (41.67 + (*px * 0.1));
    break;
    case 2:
    *py = (*px * 0.05);
    break;
    case 1:
    *py = 0;
    break;
}    
}


int entered_gross_income(float *px)

{
printf("What is your gross monthly income?: ");
scanf("%f", &*px);
if(*px<833)
    return 1;
else
    return 0;
}

void output_results(float *px, float *py, int *pc, char *pd, int *pzex)
{
printf("\nYour monthly income is %f", *px);
if (*pd == 'a')
{
    if(*pzex == 1)
        printf("\nYou're filed under Zero Exemptions");
          else
              printf("\nYou're filed under S/ME");
}
if(*pd == 'b')
   printf("\nYou're filed under ME%d/S%d", *pc, *pc);
printf("\nYour tax is %f", *py);
}

int main (void)
{
char c = 'd';
int children = 0, zex = 0, table = 0;
float income = 0, tax = 0;

if (entered_gross_income(&income)==1)
{
    printf("You have no taxes to pay!");
    exit(0);
}

print_status_menu();
entered_filing_status(&c, &children, &zex, &income, &table);
computed_tax(&income, &tax, &table);
output_results(&income, &tax, &children, &c, &zex);
}

Answer 1

条件

while (*px!='a' || *px!='b')

始终返回true。它应该是：

while(*px!='a' && *px!='b')

C：不接受字符输入+无效数学

1 个答案: