我编写了这段代码：
/*here is the main function*/
int x , y=0, returned_value;
int *p = &x;
while (y<5){
printf("Please Insert X value\n");
returned_value = scanf ("%d" , p);
validate_input(returned_value, p);
y++;
}
功能：
void validate_input(int returned_value, int *p){
getchar();
while (returned_value!=1){
printf("invalid input, Insert Integers Only\n");
getchar();
returned_value = scanf("%d", p);
}
}
虽然它通常工作得很好，但是当我插入例如“1f1”时，它接受“1”并且不报告任何错误，当插入“f1f1f”时它读取它两次并破坏第二次读取/扫描等等（即，第一次读取打印出“无效输入，仅插入整数”，而不是再次等待重新读取用户的第一次读取，而是继续第二次读取并再次打印出“无效输入，仅插入整数”。 .
它需要最后的润色，我读了很多答案但找不到它。
如果您不想接受
1f1scanfscanf而是读取整行，然后检查它是否只包含数字。之后就可以拨打
sscanf类似：
#include <stdio.h>
int validateLine(char* line)
{
int ret=0;
// Allow negative numbers
if (*line && *line == '-') line++;
// Check that remaining chars are digits
while (*line && *line != '\n')
{
if (!isdigit(*line)) return 0; // Illegal char found
ret = 1; // Remember that at least one legal digit was found
++line;
}
return ret;
}
int main(void) {
char line[256];
int i;
int x , y=0;
while (y<5)
{
printf("Please Insert X value\n");
if (fgets(line, sizeof(line), stdin)) // Read the whole line
{
if (validateLine(line)) // Check that the line is a valid number
{
// Now it should be safe to call sscanf - it shouldn't fail
// but check the return value in any case
if (1 != sscanf(line, "%d", &x))
{
printf("should never happen");
exit(1);
}
// Legal number found - break out of the "while (y<5)" loop
break;
}
else
{
printf("Illegal input %s", line);
}
}
y++;
}
if (y<5)
printf("x=%d\n", x);
else
printf("no more retries\n");
return 0;
}
输入
1f1
f1f1
-3
输出
Please Insert X value
Illegal input 1f1
Please Insert X value
Illegal input f1f1
Please Insert X value
Illegal input
Please Insert X value
x=-3
另一种方法 - 避免 scanf
您可以让您的函数计算数字，从而完全绕过 scanf。它可能看起来像：
#include <stdio.h>
int line2Int(char* line, int* x)
{
int negative = 0;
int ret=0;
int temp = 0;
if (*line && *line == '-')
{
line++;
negative = 1;
}
else if (*line && *line == '+') // If a + is to be accepted
line++; // If a + is to be accepted
while (*line && *line != '\n')
{
if (!isdigit(*line)) return 0; // Illegal char found
ret = 1;
// Update the number
temp = 10 * temp;
temp = temp + (*line - '0');
++line;
}
if (ret)
{
if (negative) temp = -temp;
*x = temp;
}
return ret;
}
int main(void) {
char line[256];
int i;
int x , y=0;
while (y<5)
{
printf("Please Insert X value\n");
if (fgets(line, sizeof(line), stdin))
{
if (line2Int(line, &x)) break; // Legal number - break out
printf("Illegal input %s", line);
}
y++;
}
if (y<5)
printf("x=%d\n", x);
else
printf("no more retries\n");
return 0;
}
一般来说，我认为你最好从输入中读取所有内容（当然，在缓冲区大小的范围内），然后然后验证输入确实是正确的格式。
在您的情况下，您使用像
f1f1fscanf(...)#include <stdio.h>
#include <string.h>
#include <stdlib.h> // Used for strtol
#define OK 0
#define NO_INPUT 1
#define TOO_LONG 2
#define NaN 3 // Not a Number (NaN)
int strIsInt(const char *ptrStr){
// Check if the string starts with a positive or negative sign
if(*ptrStr == '+' || *ptrStr == '-'){
// First character is a sign. Advance pointer position
ptrStr++;
}
// Now make sure the string (or the character after a positive/negative sign) is not null
if(*ptrStr == NULL){
return NaN;
}
while(*ptrStr != NULL){
// Check if the current character is a digit
// isdigit() returns zero for non-digit characters
if(isdigit( *ptrStr ) == 0){
// Not a digit
return NaN;
} // else, we'll increment the pointer and check the next character
ptrStr++;
}
// If we have made it this far, then we know that every character inside of the string is indeed a digit
// As such, we can go ahead and return a success response here
// (A success response, in this case, is any value other than NaN)
return 0;
}
static int getLine (char *prmpt, char *buff, size_t sz) {
int ch, extra;
// Get line with buffer overrun protection.
if (prmpt != NULL) {
printf ("%s", prmpt);
fflush (stdout);
}
if (fgets (buff, sz, stdin) == NULL)
return NO_INPUT;
// If it was too long, there'll be no newline. In that case, we flush
// to end of line so that excess doesn't affect the next call.
// (Per Chux suggestions in the comments, the "buff[0]" condition
// has been added here.)
if (buff[0] && buff[strlen(buff)-1] != '\n') {
extra = 0;
while (((ch = getchar()) != '\n') && (ch != EOF))
extra = 1;
return (extra == 1) ? TOO_LONG : OK;
}
// Otherwise remove newline and give string back to caller.
buff[strlen(buff)-1] = '\0';
return OK;
}
void validate_input(int responseCode, char *prompt, char *buffer, size_t bufferSize){
while( responseCode != OK ||
strIsInt( buffer ) == NaN )
{
printf("Invalid input.\nPlease enter integers only!\n");
fflush(stdout); /* It might be unnecessary to flush here because we'll flush STDOUT in the
getLine function anyway, but it is good practice to flush STDOUT when printing
important information. */
responseCode = getLine(prompt, buffer, bufferSize); // Read entire STDIN
}
// Finally, we know that the input is an integer
}
int main(int argc, char **argv){
char *prompt = "Please Insert X value\n";
int iResponseCode;
char cInputBuffer[100];
int x, y=0;
int *p = &x;
while(y < 5){
iResponseCode = getLine(prompt, cInputBuffer, sizeof(cInputBuffer)); // Read entire STDIN buffer
validate_input(iResponseCode, prompt, cInputBuffer, sizeof(cInputBuffer));
// Once validate_input finishes running, we should have a proper integer in our input buffer!
// Now we'll just convert it from a string to an integer, and store it in the P variable, as you
// were doing in your question.
sscanf(cInputBuffer, "%d", p);
y++;
}
}
作为免责声明/注意：我已经很长时间没有用 C 编写了，所以如果这个例子中有任何错误，我提前道歉。在发布之前我也没有机会编译和测试这段代码，因为我现在很着急。
如果您正在读取一个输入流，您知道它是一个文本流，但您不确定仅包含整数，那么请读取字符串。
此外，一旦您读取了一个字符串并想查看它是否是整数，请使用标准库转换例程
strtol()long#include <stdbool.h>
#include <stdio.h>
#include <stdlib.h>
bool convert_to_long(long *number, const char *string)
{
char *endptr;
*number = strtol(string, &endptr, 10);
/* endptr will point to the first position in the string that could
* not be converted. If this position holds the string terminator
* '\0' the conversion went well. An empty input string will also
* result in *endptr == '\0', so we have to check this too, and fail
* if this happens.
*/
if (string[0] != '\0' && *endptr == '\0')
return false; /* conversion succesful */
return true; /* problem in conversion */
}
int main(void)
{
char buffer[256];
const int max_tries = 5;
int tries = 0;
long number;
while (tries++ < max_tries) {
puts("Enter input:");
scanf("%s", buffer);
if (!convert_to_long(&number, buffer))
break; /* returns false on success */
printf("Invalid input. '%s' is not integer, %d tries left\n", buffer,
max_tries - tries);
}
if (tries > max_tries)
puts("No valid input found");
else
printf("Valid input: %ld\n", number);
return EXIT_SUCCESS;
}
添加注释：如果将
basestrtol()0x00