C语言问题：给定一个任意字符串，提取出其中所包含的整数。 c语言问题，如何将字符串中的数字提取
C\u8bed\u8a00:\u4e00\u4e2a\u7ed9\u5b9a\u7684\u5b57\u7b26\u4e32,\u63d0\u53d6\u5176\u4e2d\u7684\u6570\u5b57,\u8fde\u5728\u4e00\u8d77\u6784\u6210\u4e00\u4e2a\u6574\u6570,\u518d\u4e58\u4ee5513,\u5c31\u662f\u7ed3\u679c#include #define N 100int main() { int i; long r;//\u8981\u6ce8\u610f\u7ed3\u679c\u53ef\u80fd\u4f1a\u8d85\u51fa\u666e\u901a\u6574\u5f62\u8303\u56f4 char s[N]; scanf("%s",s); i=0; r=0; while(s[i]!='\0'){ if(s[i]>='0'&& s[i]<='9') r=r*10+(s[i]-'0'); ++i; } // r*=513; printf("\u8f93\u5165\u4e32:s=%s\n\u63d0\u53d6\u6570\u5b57:r=%ld\n\u7ed3\u679c:r*513=%ld\n",s,r,r*513); return 0;}
1\u3001\u9996\u5148\u6253\u5f00visual studio\u8f6f\u4ef6\uff0c\u65b0\u5efa\u4e00\u4e2aC\u8bed\u8a00\u6587\u4ef6\uff0c\u5982\u4e0b\u56fe\u6240\u793a\u3002
2\u3001\u63a5\u7740\u5728C\u8bed\u8a00\u6587\u4ef6\u7684\u9876\u90e8\u5bfc\u5165\u5e93\u5185\u5bb9\uff0c\u5982\u4e0b\u56fe\u6240\u793a\u3002
3\u3001\u63a5\u7740\u8fd0\u7528scanf\u51fd\u6570\u63a5\u6536\u7528\u6237\u8f93\u5165\u7684\u5b57\u7b26\u4e32\u3002
4\u3001\u7136\u540e\u5229\u7528printf\u51fd\u6570\u6253\u5370\u4e00\u4e0b\u7528\u6237\u8f93\u5165\u7684\u5185\u5bb9\u3002
5\u3001\u8fd0\u884c\u7a0b\u5e8f\u4ee5\u540e\u5c31\u4f1a\u5f39\u51fa\u5982\u4e0b\u56fe\u6240\u793a\u7684CMD\u754c\u9762\uff0c\u8f93\u5165\u5185\u5bb9\u5c31\u4f1a\u81ea\u52a8\u63a5\u6536\u5230\uff0c\u7136\u540e\u539f\u6837\u8f93\u51fa\u5185\u5bb9\u3002
6\u3001\u6700\u540e\u5982\u679c\u63a5\u6536\u5176\u4ed6\u7684\u7c7b\u578b\uff0c\u6bd4\u5982\u6574\u6570\uff0c\u90a3\u4e48\u7c7b\u578b\u5c31\u9700\u8981\u6539\u53d8\u6210int\uff0c\u5982\u4e0b\u56fe\u6240\u793a\uff0c\u5e76\u4e14scanf\u91cc\u9762\u7684\u63a5\u6536\u7c7b\u578b\u53d8\u4e3ad%\uff0c\u63d0\u53d6\u5b57\u7b26\u4e32\u4e2d\u7684\u6570\u5b57\u3002
查找字符串中的数字并返回其个数：
#include <stdio.h>
int main(void)
{
char str[1000];
int i, count, a[100];
int p, q, r;
int tmp;
gets(str);
r = 0;
count = 0;
while(1)
{
while(str[r] && (str[r]<'0' || str[r]>'9'))
r++; //跳过非数字
if(str[r])
{
p = r; //p指向数字子串开头
q = r + 1; //q寻找数字串结尾
a[count] = str[r]-'0';
while(str[q]>='0' && str[q]<='9')
{
a[count] = 10*a[count]+(str[q]-'0');
q++;
}
r = q; //新起点
count++;
}
else
break;
}
printf("共计 %d 个整数：", count);
for(i=0; i<count; i++)
printf("%d ", a[i]);
printf("
");
return 0;
}
具体代码不给你写了，很简单，就是判断是否是整数，ASCII码大于'0'小于'9'均为整数，保存下来，如果下一个是整数，继续保存，如果下一个不是整数，用空格代替，并继续读取。最后输出保存的结果。
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