1.整形转字符串

A.无符号整形数转字符串 

void Int2Str(u8* str, u32 intnum)
{u32 i, Div = 1000000000, j = 0, Status = 0;//32位无符号数最大是10位整数,所以Div=10 0000 0000for (i = 0; i < 10; i++){str[j++] = (intnum / Div) + '0';//取最高位 转化成字符 intnum = intnum % Div;//去掉最高位Div /= 10;//还剩下10-i位要转换if ((str[j-1] == '0') && (Status == 0))//忽略最高位的'0'{j = 0;}else{Status++;}}
}

u8 str[]={0};Int2Str(str,2018);
printf("%s",str);

B.有符号整形转字符串

/******************************************************************* @brief int2str** 功能：将整形(int)数据转换成字符串** @param: n = 要转换的整形数据* @param: str = 存放字符串的首地址* @param: length = 转换后字符串的长度*****************************************************************/
void int2str(int n, char *str,unsigned char *length)
{char buf[6] = "";int i = 0;int len = 0;// temp为n的绝对值int temp = n<0?-n:n;// 如果str为空，直接返回if(str == NULL){return ;}//把tmp的每一位的数存入buf// 逆序存入，如1234 -> 在buf中为4321//memset(CHAR,0,5);while(temp){buf[i++] = (temp%10) + '0';temp = temp / 10;}//如果是负数多留一位放符串len = n<0?++i:i;*length = len; str[i] = 0;//反转字符串4321- 转成 -1234while(1){i--;if(buf[len-i-1] == 0){break;}str[i] = buf[len-i-1];}// 如果是符号，加回去if(i==0){str[i] = '-';//添加负号}
}

2.字符串转整形

方法一（垃圾！！！）：

#define IS_CAP_LETTER(c)    (((c) >= 'A') && ((c) <= 'F'))
#define IS_LC_LETTER(c)     (((c) >= 'a') && ((c) <= 'f'))
#define IS_09(c)            (((c) >= '0') && ((c) <= '9'))
#define ISVALIDHEX(c)       (IS_CAP_LETTER(c) || IS_LC_LETTER(c) || IS_09(c))
#define ISVALIDDEC(c)       IS_09(c)
#define CONVERTDEC(c)       (c - '0')#define CONVERTHEX_ALPHA(c) (IS_CAP_LETTER(c) ? ((c) - 'A'+10) : ((c) - 'a'+10))
#define CONVERTHEX(c)       (IS_09(c) ? ((c) - '0') : CONVERTHEX_ALPHA(c))u32 Str2Int(u8 *p_inputstr, u32 *p_intnum)
{u32 i = 0, res = 0;u32 val = 0;if ((p_inputstr[0] == '0') && ((p_inputstr[1] == 'x') || (p_inputstr[1] == 'X'))){i = 2;while ( ( i < 11 ) && ( p_inputstr[i] != '\0' ) ){if (ISVALIDHEX(p_inputstr[i])){val = (val << 4) + CONVERTHEX(p_inputstr[i]);}else{/* Return 0, Invalid input */res = 0;break;}i++;}/* valid result */if (p_inputstr[i] == '\0'){*p_intnum = val;res = 1;}}else /* max 10-digit decimal input */{while ( ( i < 11 ) && ( res != 1 ) ){if (p_inputstr[i] == '\0'){*p_intnum = val;/* return 1 */res = 1;}else if (((p_inputstr[i] == 'k') || (p_inputstr[i] == 'K')) && (i > 0)){val = val << 10;*p_intnum = val;res = 1;}else if (((p_inputstr[i] == 'm') || (p_inputstr[i] == 'M')) && (i > 0)){val = val << 20;*p_intnum = val;res = 1;}else if (ISVALIDDEC(p_inputstr[i])){val = val * 10 + CONVERTDEC(p_inputstr[i]);}else{/* return 0, Invalid input */res = 0;break;}i++;}}return res;
}

u8 strs[]="123456";Str2Int(strs,&Num);
printf("%d\r\n",Num);

方法二（转载另一位网友的）：

这里给出原文地址：https://blog.csdn.net/second60/article/details/80503335

int str2int1( char* str)
{int temp = 0;const char* p = str;if(str == NULL) return 0;if(*str == '-' || *str == '+'){str ++;}while( *str != 0){if( *str < '0' || *str > '9'){break;}temp = temp*10 +(*str -'0');str ++;}if(*p == '-'){temp = -temp;}return temp;
}