c语言编程函数 如何在C语言中定义一个函数？
C\u8bed\u8a00\u7f16\u7a0b\uff0c\u51fd\u6570#include #include /* \u8f97\u8f6c\u76f8\u9664\u6cd5\u6c42\u6700\u5927\u516c\u7ea6\u6570 */int gcd (int m, int n) {if (n)while((m %= n) && (n %= m)); /* \u4e24\u6570\u53cd\u590d\u76f8\u9664\u53d6\u4f59\uff0c\u76f4\u81f3\u88ab\u6574\u9664 */return m+n;}int lcm (int m, int n) {return m * n / gcd (m, n);}int main (void) {int m, n;int gcdnum, lcmnum; /* \u6700\u5927\u516c\u7ea6\u6570\u3001\u6700\u5c0f\u516c\u500d\u6570 */m = 252; n = 105; /* \u6700\u5927\u516c\u7ea6\u6570\u4e3a21\uff0c\u6700\u5c0f\u516c\u500d\u6570\u4e3a1260 */gcdnum = gcd (m, n);lcmnum = lcm (m, n);printf ("%d %d\n", m, n);printf ("The gcd is %d\n", gcdnum);printf ("The lcm is %d\n", lcmnum);getch (); /*\u5c4f\u5e55\u6682\u7559*/return 0;}
\u9700\u8981\u51c6\u5907\u7684\u6750\u6599\u5206\u522b\u6709\uff1a\u7535\u8111\u3001C\u8bed\u8a00\u7f16\u8bd1\u5668\u3002
1\u3001\u9996\u5148\uff0c\u6253\u5f00C\u8bed\u8a00\u7f16\u8bd1\u5668\uff0c\u65b0\u5efa\u4e00\u4e2a\u521d\u59cb.cpp\u6587\u4ef6\uff0c\u4f8b\u5982\uff1atest.cpp\u3002
2\u3001\u5728test.cpp\u6587\u4ef6\u4e2d\uff0c\u8f93\u5165C\u8bed\u8a00\u4ee3\u7801\uff1aint fun(){return 1;}\u3002
3\u3001\u7f16\u8bd1\u5668\u8fd0\u884ctest.cpp\u6587\u4ef6\uff0c\u6b64\u65f6\u6210\u529f\u8c03\u7528fun()\u51fd\u6570\u8f93\u51fa\u4e86\u5185\u5bb9\u3002
*in: 需要测长度的字符串
*out: 字符串长度
*/
int strlen(string str)
{
assert( str != NULL ); // #include <assert.h>
char *ch = NULL;
int iLength = 0;
ch = str;
while ( *ch != '\0')
{
ch = ch + 1;
iLength = iLength + 1;
}
return iLength;
}
#include <stdio.h>
int Strlen(char *str);
int main(int argc, char *argv[])
{
char *tmp = "faadfj";
printf("strlen = %d\n", Strlen(tmp));
return 0;
}
int Strlen(char *str)
{
if( !str )
return 0;
int len = 0;
char *tmp = str;
while ( *tmp!='\0')
{
tmp++;
len++;
}
return len;
}
int mystrlen(char *str)
{
char *tmp=str;
while(*str++);
return str-tmp-1;
}
int strlen(char* str)
{
register int a = 0;
while(*str++)a++;
return a;
}
int
slen(s)
char *s;
{
register char *p = s;
while(*p)
p++;
return(p - s);
}
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