An algorithm designed to deal with computational equations

For example, when we do a calculator we will encounter 1+9+6/8 similar to this, then we can use this formula to become a suffix expression and then to calculate.

struct//set operator Precedence {char ch;//operator int pri;//priority}lpri[] = {' = ', 0}, {' (', 1}, {' * ', 5}, {'/', 5}, {' + ', 3}, {' -', 3}, {') ', 6}},rpri[] = {{' = ', 0}, {' (', 6}, {' * ', 4}, {' + ', 4}, {'-', 2}, {') ', 2}};int Leftpri (char op)//The priority of the left operator {int i;for (i = 0; i<7; i++) if (lpri[i].ch = = op) return lpri[i].pri;} int Rightpri (char op)//right operator precedence {int i;for (i = 0; i<7; i++) if (rpri[i].ch = = op) return rpri[i].pri;} BOOL Inop (char ch)//judgment CH is not an operator {if (ch = = ' (' | | ch = = ') ' | | ch = = ' + ' | | ch = = '-' | | | ch = = ' * ' | | ; Elsereturn false;} The precedence condition of the int precede (char OP1, char op2)//OP1 and OP2 operators {if (LEFTPRI (OP1) = = Rightpri (OP2)) return 0;if (Leftpri (OP1) < Rightpri (OP2)) Return-1;elsereturn 1;} void Trans (char *exp, char postexp[])//convert general expression to suffix expression {struct//store operator {char data[1000];int top;//stack pointer}op;int i = 0;op.top = -1;op.top++;op.data[op.top] = ' = '; while (*exp! = ')//exp expression is not scanned when the loop {if (! Inop (*EXP))//If the number {while (*exp >= ' 0 ' &&*exp <= ' 9 ')//determine the number {if (* (exp-2) = = ' (' && (* (exp-1) = = ' + ' | | * (exp-1) = = '-' | | * (exp-1) = = ' * ' | | * (EXP- 1) = = '/')//is a negative sign instead of a minus sign {postexp[i++] = ' & ';//This is to deal with negative cases, if it is a negative number will appear & and numbers such as &5 int a = i + 1;postexp[a++] = *exp;}  postexp[i++] = *exp;exp++;//store expression}postexp[i++] = ' # ';//Use a # number to separate}else//if it is an operator {if (*exp = = ' + ') || *exp = = '-' | | *exp = = ' * ' | | *exp = = '/') && * (exp-1) = = ' (' && (* (exp + 1) >= ' 0 ' &&* (exp + 1) <= ' 9 ')//get rid of negative number-No. {exp+ +;continue;} Else{switch (Precede (op.data[op.top], *exp)) {case-1://stack top operator low priority op.top++;op.data[op.top] = *exp;exp++;break;case 0 ://Only parentheses satisfy this situation op.top--;exp++;break;case 1://and enter into Postexp array postexp[i++] = Op.data[op.top];op.top--;break;}}} while (op.data[op.top]! = ' = ')//At this time has been scanned, the stack to ' = ' End {postexp[i++] = op.data[op.top];op.top--;} Postexp[i] = ' + ';//Add an End flag}float compvalue (char *postexp)//start with a suffix expression to calculate the value {struct{float data[1000];//the array int top that holds the number;} St;float D, A, B, c;st. top = -1;while (*postexp! =)//When the Postexp string is scanned {switch (*POSTEXP) {case ' + '://Judgment is ' + ' when a = st.data[st.top];st.top-- b = st.data[st.top];st.top--;c = a + b;st.top++;st.data[st.top] = c;//This is the calculation of the sum of two numbers when a, a, a, a, and then the stack break;case '-': a = St.da Ta[st.top];st.top--;b = St.data[st.top];st.top--;c = b-a;st.top++;st.data[st.top] = C;break;case ' * ': a = St.data[st.top ];st.top--;b = St.data[st.top];st.top--;c = b*a;st.top++;st.data[st.top] = C;break;case '/': a = st.data[st.top]; St.top--;b = St.data[st.top];st.top--;if (A! = 0) {c = b/a;st.top++;st.data[st.top] = c;} else{printf ("error (denominator 0) \ n"); exit (0);} Break;default:d = 0;int find = 0;while (*postexp >= ' 0 ' &&*postexp <= ' 9 ' &&*postexp! = ' & ') {if (* (postexp-1) = = ' & ') {find = 1;} Else{find =-1;} D = Ten * d + *postexp-' 0 ';p ostexp++;} if (find = = 1) {St.top++;st.data[st.top] =-D;} else if (find = =-1) {st.top++;st.data[st.top] = D;} break;} postexp++;} Return (St.data[st.top]);//The position of the last pointer refers to the value}

This is a C + + version, is now making this into a Java package after the dedicated use

An algorithm designed to deal with computational equations