检查输入到char数组中的所有值是否均为数字 [英] Check if all values entered into char array are numerical
问题描述
我有一个char数组,其中只能是数字(10位数字).如果用户输入字母或特殊字符(甚至在这些数字中),程序应再次提示用户输入数字.
I have a char array, where should be only numbers (10 digits). If the user enters letters or special characters (even among those digits) the program should prompt the user again to enter the number.
我尝试了很多方法,但是仍然找不到方法.到目前为止,这是我可以做的:
I tried so many ways of doing it, but still couldn't find a way. That's what I could do so far:
int f = 1;
int i = 0;
int flag =1;
char num[11];
printf("Enter a number: ");
while (f == 1) {
scanf("%10s", num);
while (flag == 1 && isdigit(num[i])) {
i++;
if (i == 10) {
f = 0;
flag =0;
}
}
if (!isdigit(num[i])) {
printf("Enter numerical char: ");
}
}
输入错误值后,它会显示一个空行.如果我在该空行中输入了任何值,则只会显示输入数字字符:".并提示再次输入数字.
After I enter incorrect value it displays an empty line. If I put any value to that empty line, only then it says "Enter numerical char: " and prompts to enter the num again.
P.S.考虑到与数字不同的字符(不包括数字),我知道有一种方法可以确保仅输入数值.(不知道如何执行此操作)
P.S. I know there is a way of ensuring that only numerical values entered, considering the fact that characters, unlike numbers are included in single quotes. (have no idea how to do this tbh)
提前谢谢!
推荐答案
有两种方法可以做到这一点.
There are a couple of ways to do this.
简单的方法是使用 fgets
获取字符串.
The simple way is to use fgets
to get a string.
然后,我们可以使用 strtol
解码数字[并检查结尾字符的有效性].
Then, we can use strtol
to decode the number [and we check the ending char for validity].
要完全手动完成此操作,我们可以循环使用 isdigit
,一次建立一个数字.
To do it completely manually, we can use isdigit
in a loop, building up the number one digit at a time.
下面是带注释的示例代码,它们显示了两种方式:
Here's some example code that is annotated and shows both ways:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <errno.h>
#include <ctype.h>
// getstr -- get a string with prompt
// RETURNS: length or (<0 -> error)
int
getstr(char *buf,int buflen,const char *prompt)
{
char *cp;
int ret = 0;
// NOTE: usage of the error codes in errno.h is arbitrary
while (ret <= 0) {
// ensure buffer has enough space
if (buflen < 2) {
ret = -ENOMEM;
break;
}
// output prompt
printf("%s: ",prompt);
fflush(stdout);
// get a line
cp = fgets(buf,buflen,stdin);
// EOF
if (cp == NULL) {
ret = -ENODATA;
break;
}
// get buffer length
ret = strlen(buf);
// empty string
if (ret <= 0)
continue;
// point to last char
cp = &buf[ret - 1];
// ensure we got a newline -- if not, fgets had to chop the line (i.e.)
// the line is too long to fit in the buffer
if (*cp != '\n') {
ret = -ENOSPC;
break;
}
// strip the newline -- we are done
*cp = 0;
--ret;
}
return ret;
}
// getnum_strtol -- get number using strtol
long
getnum_strtol(const char *prompt)
{
int len;
int readflg = 1;
char *cp;
char buf[100];
long num = 0;
while (readflg) {
len = getstr(buf,sizeof(buf),prompt);
if (len < 0)
exit(1);
num = strtol(buf,&cp,10);
// ensure we got a least one digit
if (cp <= buf)
continue;
switch (*cp) {
case ' ':
case '\t':
case 0:
readflg = 0;
break;
default:
printf("getnum_strtol: not a valid number -- buffer '%s', invalid '%s'\n",
buf,cp);
break;
}
}
return num;
}
// getnum_manual -- get number _not_ using strtol
long
getnum_manual(const char *prompt)
{
int len;
int readflg = 1;
int sign = 0;
int valid;
int chr;
char *cp;
char buf[100];
long num = 0;
while (readflg) {
len = getstr(buf,sizeof(buf),prompt);
// fatal error
if (len < 0)
exit(1);
// point to buffer start
cp = buf;
// find first non-whitespace character
valid = 0;
while (1) {
chr = *cp;
// end of string
if (chr == 0)
break;
// found character
valid = ((chr != ' ') && (chr != '\t'));
if (valid)
break;
++cp;
}
if (! valid)
continue;
// reset the accumlated number and the sign
num = 0;
sign = 0;
valid = 0;
// loop through all characters in buffer
while (1) {
chr = *cp++;
// get the sign of the number (and skip an explicit sign)
if (sign == 0) {
switch (chr) {
case '+':
sign = 1;
chr = *cp++;
break;
case '-':
sign = -1;
chr = *cp++;
break;
default:
sign = 1;
break;
}
}
// stop decoding number on whitespace
switch (chr) {
case ' ':
case '\t':
chr = 0;
break;
}
// check for clean end of number
if (chr == 0) {
if (valid) {
readflg = 0;
break;
}
}
// not a valid digit
if (! isdigit((unsigned char) chr)) {
cp -= 1;
printf("getnum_manual: not a valid number -- buffer '%s', invalid '%s'\n",
buf,cp);
break;
}
// add digit to number
num *= 10;
chr -= '0';
num += chr;
// we got at least one valid digit
valid = 1;
}
}
// apply sign
num *= sign;
return num;
}
int
main(int argc,char **argv)
{
char *cp;
int opt_s = 0;
long num;
// skip over program name
--argc;
++argv;
// get options
for (; argc > 0; --argc, ++argv) {
cp = *argv;
if (*cp != '-')
break;
cp += 2;
switch (cp[-1]) {
case 's': // use strtol
opt_s = ! opt_s;
break;
}
}
while (1) {
if (opt_s)
num = getnum_strtol("Enter number [strtol]");
else
num = getnum_manual("Enter number [manual]");
printf("The number entered is: %ld\n",num);
if (num == 999)
break;
}
return 0;
}
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