str*函数和大小端判断

#include <stdio.h>
#include <assert.h>

size_t  mstrlen(const char *s)
{
	assert(s != NULL);

	if (s == NULL) {
		return 0;
	}

	size_t ret = 0;

	while (*s != '') {
		++ret;
		++s;
	}

	return ret;
}

void  test_strlen()
{
	char a[] = "hello,world";

	printf("%lu
", mstrlen(a));

}


char *mstrcpy(char *dst, const char *src)
{
	assert(src != NULL);//若条件不满足(为假),则终止该函数;
	assert(dst != NULL);//同上.

	char *pdst = dst;
//当函数执行到字符串的末尾时‘ ’,条件不满足则结束函数  返回“ pdst”.
	while (*dst = *src) {
	 	++dst;
		++src;
	 }
	/* RTFSD: read the fuck source code */
	/*  hello,world'' = '' = 0*/

	return pdst;//最后返回的值,结果为指针dst的首地址;
}

void  test_strcpy()  //??????
{
	char d[4] = {0};
	char s[] = "hello";
    mstrcpy(d,s);
	printf("strcpy:d:%s
", mstrcpy(d, s));//结果为strcpy:d:hello
   printf("%c
",d[5]);//结果为何还为“"o"  ?????
}

char *mstrncpy(char *dest, const char *src, size_t n)
{
	assert(dest != NULL);
	assert(src != NULL);

	char *pdest = dest;
//将指针src所指的内容一步步复制给指针dest所指的地址内;
	while (n--) {
		if (*src != '') {
			*dest++ = *src++;
		}else {
			*dest++ = '';//字符串的结束标志复制给指针dest所指的地址,此时复制也就结束了;
		}
	}

	return pdest;//返回指针dest的首地址;
}

void test_strncpy()
{
	char d[30] = "abcdefghijk";
	char s[10] = "hello";
    mstrncpy(d,s,8);
	printf("strncpy:%s
", mstrncpy(d, s, 8));//结果为strncpy:hello
    printf("%c,%c,%c
",d[6],d[7],d[11]);//结果为空白,因为复制过去的没有内容。。
}

/*将*src的前n个字符赋给*dest的末尾*/
char *mstrncat(char *dest, const char *src, size_t n)
{
	assert(dest != NULL);
	assert(src != NULL);

	char *pdest = dest;

while (*dest != '') {
		++dest;//指针所指的地址依次移动;
	}

	while (n--) {
		if (*src != '') {
			*dest++ = *src++;//将指针src所指地址的值赋给指针dest所指的地址,同时向右移动一个地址;
		}else {
			*dest++ = '';
		}
	}

	return pdest;
}
/*将*src的前n个字节的内容拷贝给*dest,不会因为字符串的结束而结束*/
void *mmemcpy(void *dest, const void *src, size_t n)
{
	char *pdest = (char *)dest;
	const char *psrc = (char *)src;

	while (n--) {/*先判断n,再进行自减*/
		*pdest++ = *psrc++;
	}

	return dest;
}


void test_memcpy()
{
	char a[] = {0x11, 0x22, 0x33, 0x44, 0x55};

	int b = 0;
	mmemcpy(&b, a, sizeof(b));

	printf("b = %#x", b);//结果为0x44332211,按照小端模式排列
    printf("b = %x", b);//结果为44332211,
}

/*  s:hello,world  accpet: abcde *///在字符串*s中找到最先含有*accept的任一字符的位置并返回,若没有则返回空指针
char *mstrpbrk(const char *s, const char *accept)
{
	assert(s != NULL);
	assert(accept != NULL);

	const char *t = NULL;//将空指针赋值给*t;
//先在对s进行一层循环 ,后在对accept进行一层循环;
	for (; *s != ''; ++s) {
			for (t = accept; *t  != ''; ++t) {
				if (*s == *t) {
					return (char *)s;//找到了 就将他的地址返回出来,不再执行后面的语句了;
				}
			}
	}

	return NULL;//没找到,此时返回空指针;
}
void test_strpbrk(void)
{
	char s[] = "hello,world";
	char a[] = "abcde";

	printf("found = %c
", *mstrpbrk(s, a));//结果为found=e;
}
//在字符串haystack中查找字符串needle 若找到就返回出来haystack找到的字符串后所以内容,  若没有则返回空指针;
char *mstrstr(const char *haystack, const char *needle)
{
	assert(haystack != NULL);
	assert(needle != NULL);

	const char *ph = NULL;
	const char *pn = NULL;


	for (; *haystack != ''; ++haystack)  // '0' != ''  == 0
	{
		for (ph = haystack, pn = needle; (*pn != '') &&
				(*ph == *pn); ++ph, ++pn) {
			;
		}
		/* pn 到达字符结尾*/	
		if (*pn == '') {
			return (char *)haystack;//返回此时的haystack及后面所跟的所有内容;
		}
	}
		
		return NULL;
}

void test_strstr()
{
	char a[] = "hello,worldxwordyfff";
	char b[] = "word";
	
	printf("strstr:%s
", mstrstr(a, b));//strstr:wordyfff

}
/*字符串反转*/
char *strreverse(char *s)
{
	assert(s != NULL);//断言
	
	char *head = s;
	char *tail = s;
	char c = 0;

	while (*(tail+1) != '') {
		++tail;//将指针tail的地址移动到最高位;
	}
	
	for (; head < tail; ++head, --tail) {
		c = *head;/*将指针head指的内容和指针tail指向的内容进行交换*/
		*head = *tail;
		*tail = c;
	}
	
	return s;//为何要返回s???
}

void test_strreverse()
{
	char a[] = "hello,world";

	printf("strreverse:%s
", strreverse(a));//结果为strreverse:dlrow,olleh

}
// "1234" --> 1234 /*将字符型转换成整型*/
int matoi(const char *s)
{
	if (s == NULL) {
		return 0;/*若为空指针则返回0,并跳出这个函数*/
	}
	
	int ret = 0;
	
	int n = 0;

	while (*s != '') {
		n = *s - '0';/*将字符型转化成整形*/
		ret = ret * 10 + n;
		++s;/*向高位自行移动一位地址*/
	}

	return ret;/*返回得到的整型数*/
}

void test_atoi()
{
	printf("atoi:%d
", matoi("1234567"));
}

/* num = 1234567, buf="1234567" *//*将整型转换成字符型,buf是转换的指针,num被转换成的数组,base转换进制类型*/
char *mitoa(char *buf, int num, int base)
{
	assert(buf != NULL);

	char tab[] = "0123456789abcdef";
	
	char *pb = buf;
	
	while (num != 0) {
		*pb++ = tab[num % base];/*不断的除进制数进行取余*/
		num /= base;//等价于num=num/baes
	}
	*pa=;
#if DEBUG
	printf("buf=%s
", buf);
#endif
	return strreverse(buf);//此处是将buf里面的内容进行反转
}

void test_itoa()
{
	char buf[20] = {0};
	mitoa(buf, 0xf0, 2);
	printf("itoa:%s
", buf);//结果为itoa:11110000  (如何输出检测到停止位)?
}



/*8月6号的任务题目*/
// fmt: "abc%co%dm%x%s%%"//  ???没弄懂
int msprintf(char *buf, const char *fmt, ...)
{
	int ret = 0;
	va_list ap;//你的类型列表
	va_start(ap, fmt);//初始化你的ap链表
	
	enum {OFF, ON} flag = OFF;//OFF=0;ON=1
	
	//printf("sizeof(enum)%lu
", sizeof flag);
	
	for (; *fmt != ''; ++fmt) {
		switch (*fmt) {
			case '%':
					if (flag == OFF) {
						//格式位
						flag = ON;
					}else {
						//普通字符
						flag = OFF;
						*buf++ = *fmt;
						++ret;
					}
					break;
			case 'c':
					if (flag == OFF) {
						*buf++ = *fmt;
					}else{
						flag = OFF;
						*buf++ = (char)va_arg(ap, int);
					}		
						++ret;
					break;
			case 'd':
					if (flag == OFF) {
						*buf++ = *fmt;
						++ret;
					}else {
						flag = OFF;
						mitoa(buf, va_arg(ap, int), 10); // "1234"
						while (*buf != '') {
							++buf;
							++ret;
						}
					}
					break;
			case 'o':
					if (flag == OFF) {
						*buf++ = *fmt;
						++ret;
					}else {
						flag = OFF;
						mitoa(buf, va_arg(ap, int), 8); // "1234"
						while (*buf != '') {
							++buf;
							++ret;
						}
					}
					break;
			case 'x':
					if (flag == OFF) {
						*buf++ = *fmt;
						++ret;
					}else {
						flag = OFF;
						mitoa(buf, va_arg(ap, int), 16); // "1234"
						while (*buf != '') {
							++buf;
							++ret;
						}
					}
					break;
			case 's':
					if (flag == OFF) {
						*buf++ = *fmt;
						++ret;
					} else {
						flag = OFF;
						mstrcpy(buf, va_arg(ap, char *));
						while (*buf != '') {
							++buf;
							++ret;
						}
					}
					break;
			default:
					*buf++ = *fmt;
					++ret;
					break;
		}
	}

	va_end(ap);

	return ret;
}

void test_sprintf()
{
	char buf[30] = {0};

	int n = msprintf(buf, "a%so%dab%c%%%ox%x","hello",1234,'X',
		0127, 0x18f);

	printf("n = %d, buf = %s
", n, buf);//n=22,buf=ahelloo1234abX%127x18f
}

int main()
{
//	test_strlen();
//	test_strcpy();
.写一个函数判断系统是大端Big_endian还是小端Little_endian
大端格式:字数据的高字节存储在低地址中,而字数据的低字节存放在高地址中
小端格式:字数据的高字节存放在高地址中,而字数据的低字节存放在低地址中

大小端存储问题,如果小端方式中(i占至少两个字节的长度)则i所分配的内存最小地址那个字节中就存着1,
其他字节是0.大端的话则1在i的最高地址字节处存放,char是一个字节,所以强制将char型量p指向i则p指向
的一定是i的最低地址,那么就可以判断p中的值是不是1来确定是不是小端。
int i = 1;
char *p = (char*)&i;
if(*p == 1)
	printf("Little_endian");
else
	printf("Big_endian");
// test_strncpy(); // test_memcpy(); // test_strpbrk(); // test_strstr(); // test_strreverse(); // test_atoi(); test_itoa(); test_sprintf(); return 0; }
原文地址:https://www.cnblogs.com/0822vaj/p/3555251.html