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即時通訊有問題的scanf和獲取。我知道它的錯誤，但我找不到任何其他方式。這樣，該名稱正在打印出來，但它不打印出它的第一個字母。 這裏是我的代碼：scanf並獲取緩衝區
#include <stdio.h>
float calculations(int age, float highBP, float lowBP);
char option;
int counter, age;
char temp_name[50];
float highBP, lowBP, riskF, optimalH = 120.0, optimalL = 80.0;
typedef struct {
char name[50]; /*which represents the patient’s name*/
int age; /*which represents the patient’s age*/
float highBP; /*highBP, which represents the patient’s high (systolic) blood pressure*/
float lowBP; /*lowBP, which represents the patient’s low (diastolic) blood pressure*/
float riskF; /*riskFactor, which represents the patient’s risk factor for stroke due to hypertension.*/
}patient;/*end structure patient*/
patient *pRecords[30];
void printMenu()
{
printf("\n---------------------------------------------------------\n");
printf("|\t(N)ew record\t(D)isplay db\t(U)pdate record\t|\n");
printf("|\t(L)oad disk\t(W)rite disk\t(E)mpty disk\t|\n");
printf("|\t(S)ort db\t(C)lear db\t(Q)uit \t\t|\n");
printf("---------------------------------------------------------\n");
printf("choose one:");
}/*end print menu*/
void enter()
{
if(counter == 30)
printf("database full.");
else{
printf("name: ");
while(getchar()=='\n');
gets(temp_name);
strcpy(pRecords[counter]->name , temp_name);
printf("age: "); scanf("%d", &age);
pRecords[counter]->age = age;
printf("highBP: "); scanf("%f", &highBP);
pRecords[counter]->highBP = highBP;
printf("lowBP: "); scanf("%f", &lowBP);
pRecords[counter]->lowBP = lowBP;
float temp = calculations(age, highBP,lowBP);
pRecords[counter]->riskF = temp;
/*printf("name: %s, age: %d, highbp:%.1f, lowBP:%.1f\n", pRecords[counter]->name,pRecords[counter]->age,pRecords[counter]->highBP,pRecords[counter]->lowBP);
printf("risk factor: %.1f\n", pRecords[counter]->riskF);*/
counter ++;
}
}/*end of void enter function*/
memallocate(int counter){
pRecords[counter] = (patient *)malloc (sizeof(patient));
}/*end memallocate function*/
void display()
{
printf("===============================\n");
int i;
for(i=0; i<counter; i++)
{
printf("name: %s\n", pRecords[i]->name);
printf("age: %d\n", pRecords[i]->age);
printf("bp: %.2f %.2f\n", pRecords[i]->highBP, pRecords[i]->lowBP);
printf("risk: %.2f\n\n", pRecords[i]->riskF);
}/*end of for loop*/
printf("========== %d records ==========", counter);
}/*end of display method*/
float calculations(int age, float highBP, float lowBP)
{ float risk;
if((highBP <= optimalH) && (lowBP <= optimalL))
{ risk = 0.0;
if(age >=50)
risk = 0.5;
}
else if(highBP <= optimalH && (lowBP>optimalL && lowBP <=(optimalL+10)))
{ risk= 1.0;
if(age >=50)
risk = 1.5;
}
else if ((highBP >optimalH && highBP <= (optimalH+10))&& lowBP <=optimalL)
{ risk= 1.0;
if(age >=50)
risk= 1.5;
}
else if((highBP > optimalH && highBP <=(optimalH+10)) && (lowBP >optimalL && lowBP <= (optimalL+10)))
{ risk= 2.0;
if(age >=50)
risk = 2.5;
}
else if(highBP < optimalH && (lowBP >(optimalL+11) && lowBP<(optimalL+20)))
{ risk = 3.0;
if(age >=50)
risk = 3.5;
}
else if((lowBP < optimalL) && (highBP >(optimalH+11) && highBP<(optimalH+20)))
{ risk = 3.0;
if(age >=50)
risk = 3.5;
}
else if((highBP>=(optimalH+11) && highBP <= (optimalH+20))&& (lowBP>=(optimalL+11) && lowBP<=(optimalL+20)))
{ risk = 4.0;
if(age >=50)
risk = 4.5;
}
else
{ risk = 5.0;
if(age >=50)
risk = 5.5;
}
return risk;
}/*end of calculation function*/
main()
{
printMenu();
char option=getchar();
while(option != 'q' || option != 'Q'){
if(option == 'N' || option == 'n')
{
memallocate(counter);
enter();
printMenu();
}
if (option == 'L' || option == 'l')
{
printMenu();
}
if(option == 'S' || option == 's')
{
printMenu();
}
if(option == 'D' || option == 'd')
{
display();
printMenu();
}
if(option == 'W' || option == 'w')
{
printMenu();
}
if(option == 'C' || option == 'c')
{
printMenu();
}
if(option == 'U' || option == 'u')
{
printMenu();
}
if(option == 'E' || option == 'e')
{
printMenu();
}
if(option == 'Q' || option == 'q')
{
exit(0);
}
option = getchar();
}/*end while*/
system("pause");
}/*end main*/
輸出樣本：
---------------------------------------------------------
| (N)ew record (D)isplay db (U)pdate record |
| (L)oad disk (W)rite disk (E)mpty disk |
| (S)ort db (C)lear db (Q)uit |
---------------------------------------------------------
choose one: n
name: judy
age: 30
high bp: 110
low bp: 88
3
---------------------------------------------------------
| (N)ew record (D)isplay db (U)pdate record |
| (L)oad disk (W)rite disk (E)mpty disk |
| (S)ort db (C)lear db (Q)uit |
---------------------------------------------------------
choose one: n
name: cindy white
age: 52
high bp: 100.7
low bp: 89.4
---------------------------------------------------------
| (N)ew record (D)isplay db (U)pdate record |
| (L)oad disk (W)rite disk (E)mpty disk |
| (S)ort db (C)lear db (Q)uit |
---------------------------------------------------------
choose one: d
===============================
name: udy
age: 30
bp: 110.00 88.00
risk: 1.0
name: indy white
age: 52
bp: 100.70 89.40
risk: 1.5
========== 2 records ==========
請歸結示例代碼的測試用例，可以說明你的問題。 – jbruni