Output result string after numbers addition and subtraction

Title Description:By typing the formula for adding and reducing the number of positive integers within 100, write a program output result string. The input string is in the following format: operand 1 operator operand 2, operand and operator separated by a space.
Additional notes:1. The operand is a positive integer, and there is no need to consider the case of a calculated result overflow. 2. If the input calculation format is wrong, the output is "0".
Required Implementation functions:void arithmetic (const char *PINPUTSTR, long Linputlen, Char *poutputstr);
input pinputstr:Input String Linputlen:Input string length Outputpoutputstr: Output string, space has been opened up, and input string and so long;

<span style= "font-size:10px;" > #include <stdio.h> #include <stdlib.h> #include <string.h> #define Maxchar 10void Arithmetic (    const char *PINPUTSTR, long Linputlen, char *poutputstr) {int I, CNT = 0, A, b, result;    Char ch[1] = {' 0 '};    Char Op1[maxchar], Op[maxchar], Op2[maxchar], buffer[4];        for (i = 0; i < Linputlen; i++) if (pinputstr[i] = = ") cnt++;        if (cnt! = 2)//space number is not equal to 2 {strcat (POUTPUTSTR, ch);    Return    } sscanf (Pinputstr, "%s%s%s", OP1, OP, OP2); if (strlen (OP) > 1 | |            (op[0]! = ' + ' && op[0]! = '-'))        {strcat (POUTPUTSTR, ch);    Return } for (i = 0; i < strlen (OP1); i++) {if (Op1[i] < ' 0 ' | | op1[i] > ' 9 '            ) {strcat (POUTPUTSTR, ch);        Return }} for (i = 0; i < strlen (OP2); i++) {if (Op2[i] < ' 0 ' | |            Op2[i] > ' 9 ') {strcat (POUTPUTSTR, ch);        Return    }} A = Atoi (OP1);        b = atoi (OP2);            Switch (op[0]) {case ' + ': result = a + B;            Itoa (result, buffer, 10);            strcat (poutputstr, buffer);        Break            Case '-': result = a-B;            Itoa (result, buffer, 10);            strcat (poutputstr, buffer);        Break    Default:break;      }}int Main () {char pinputstr3[] = {"3 + 4"};    Char Poutputstr3[maxchar] = {0};      Arithmetic (PINPUTSTR3, strlen (PINPUTSTR3), POUTPUTSTR3);     printf (POUTPUTSTR3); return;} </span>

Knowledge points1.sscanf ()-Reads data in a string that matches the specified format. function Prototypes:
int sscanf (String str, String fmt, mixed var1, mixed var2 ...);
int scanf (const char *format [, argument] ...);
Description:
sscanf is similar to scanf, which is used for input, except that the latter takes a screen (stdin) as the input source, the former with a fixed string as the input source.
Reference:http://www.cnblogs.com/lyq105/archive/2009/11/28/1612677.html Reference:2.itoa ()-function Prototype: char *itoa (int value, char *string,int radix);
    Description:          Value: The data that you want to convert.
< Span style= "FONT-FAMILY:ARIAL,SANS-SERIF; line-height:28px ">       string: The address of the target string.
< Span style= "White-space:pre" >        radix: converted binary number, It can be 10 binary, 16 binary and so on.
    Example:

converts an integer to string usage itoa (i,num,10);

I----numbers that need to be converted into strings

num----A variable that saves a string after conversion

----Converts the cardinality of the number (that is, the binary). 10 means that the conversion is done by 10 binary . It could be 2,8,16 and so on. Your preferred type of production

Return value: A pointer to the string num

3.atoi () Use the opposite of iota ()

Reference Program: http://blog.csdn.net/poinsettia/article/details/9569987 Program:

Output result string after numbers addition and subtraction