C language multiple operation string, output (not easy Ah, adjusted for 1 hours, just adjust it out ....) I hope the gods have a better way of pointing twos.

Enter a multi-operand string, output the result, support only arithmetic, and support only integers.

Parsing string operations (without parentheses)

int calculatorfinal (char *str)

{

JIEGUO[100][10] for storing parsed arrays

Char *p[100], jieguo[100][10];

int result, secondnumber, i = 0, j = 0;

Char tempnumber[20], op;

for (i = 0; i <; i++) {

P[i] = Jieguo[i];

}

i = 0;

Parse the string and put it into a two-dimensional array

while (*STR) {

if (*str >= ' 0 ' && *str <= ' 9 ') {

TEMPNUMBER[J] = *str;

j + +;

}else{

TEMPNUMBER[J] = ' + ';

strcpy (P[i], tempnumber);

i++;

for (int j = 0; J <; J + +) {

TEMPNUMBER[J] = ' + ';

}

Tempnumber[0] = *str;

strcpy (P[i], tempnumber);

i++;

Tempnumber[0] = ' + ';

j = 0;

}

str++;

}

strcpy (P[i], tempnumber);

i++;

for (int j = 0; J <; J + +) {

TEMPNUMBER[J] = ' + ';

}

Tempnumber[0] = *str;

strcpy (P[i], tempnumber);

printf ("* * * * parsing string result ********\n");

for (j = 0; J < i; J + +) {

Puts (P[j]);

//    }

printf ("* * * * * multiplication law ********\n");

for (int n = 0; n < i; n++) {

if (n% 2 = = 1) {

op = p[n][0];

if (op = = ' * ') {

result = Atoi (p[n-1]) * ATOI (p[n + 1]);

printf ("result =%d\n", result);

for (int m = n; m < i; m++) {

if (M = = N) {

printf ("result =%d\n", result);

sprintf (P[m-1], "%d", result);

Puts (p[m-1]);

}

To determine if this calculation is the last group, the words stop moving the array, not the words move

if (M + 2! = i) {

strcpy (P[m], p[m + 2]);

strcpy (p[m + 1], P[m + 3]);

}else if (m + 2 = = i) {

Each time an operation is performed, the array moves up and starts to detect ' * ', '/'

and reduce the array size by 2

n = 0;

I-= 2;

Break

}

}

}else if (op = = '/') {

result = Atoi (p[n-1])/atoi (P[n + 1]);

for (int m = n; m < i; m++) {

if (M = = N) {

printf ("result =%d\n", result);

sprintf (P[m-1], "%d", result);

Puts (p[m-1]);

}

To determine if this calculation is the last group, the words stop moving the array, not the words move

if (M + 2! = i) {

strcpy (P[m], p[m + 2]);

strcpy (p[m + 1], P[m + 3]);

}else if (m + 2 = = i) {

Each time an operation is performed, the array moves up and starts to detect ' * ', '/'

and reduce the array size by 2

n = 0;

I-= 2;

Break

}

}

}

}

}

printf ("* * * * * * multiplication----results ********\n");

for (j = 0; J < i; J + +) {

Puts (P[j]);

//    }

printf ("* * * * Add subtraction----results ********\n");

for (int n = 0; n < i; n++) {

if (n% 2 = = 1) {

op = p[n][0];

if (op = = ' + ') {

result = Atoi (p[n-1]) + atoi (p[n + 1]);

printf ("result =%d\n", result);

for (int m = n; m < i; m++) {

if (M = = N) {

printf ("result =%d\n", result);

sprintf (P[m-1], "%d", result);

Puts (p[m-1]);

}

To determine if this calculation is the last group, the words stop moving the array, not the words move

if (M + 2! = i) {

strcpy (P[m], p[m + 2]);

strcpy (p[m + 1], P[m + 3]);

}else if (m + 2 = = i) {

Each time an operation is performed, the array moves up and starts to detect ' * ', '/'

and reduce the array size by 2

n = 0;

I-= 2;

Break

}

}

}else if (op = = '-') {

result = Atoi (p[n-1])-atoi (P[n + 1]);

for (int m = n; m < i; m++) {

if (M = = N) {

printf ("result =%d\n", result);

sprintf (P[m-1], "%d", result);

Puts (p[m-1]);

}

To determine if this calculation is the last group, the words stop moving the array, not the words move

if (M + 2! = i) {

strcpy (P[m], p[m + 2]);

strcpy (p[m + 1], P[m + 3]);

}else if (m + 2 = = i) {

Each time an operation is performed, the array moves up and starts to detect ' * ', '/'

and reduce the array size by 2

n = 0;

I-= 2;

Break

}

}

}

}

}

return result;

}

C language multiple operation string, output (not easy Ah, adjusted for 1 hours, just adjust it out ....) I hope the gods have a better way of pointing twos.