String designation in C?

Question

I am working on a terminal parser for a calculator written in C. I cannot figure out how to combine all the numbers between the operators to put them in an array.

For example, if the input (command line argument) was " 4+342 ", ideally it would be input[] = {"4", "+", "342"} .

Here is my code so far. I include <stdio.h> , <stdlib.h> and <ctype.h> .

 typedef char * string; int main(int argc, char *argv[]) { string inputS = argv[1]; string input[10]; string temp; printf("%s\n", inputS); int i; int len = strlen(inputS); printf("parsed:\n"); for(i = 0; i < len; inputS++, i++) { if(isdigit(*inputS)) { printf("%c",*inputS); } else { printf("\n%c\n",*inputS); } } printf("\n"); return 0; } 

If it is running with ./calc 4+5-546 , it will output:

 4 + 5 - 546 

So, what is the easiest way to get each line of this in your own slot?

Answer 1

Try this for size ...

 #include <stdio.h> #include <ctype.h> typedef char * string; int main(int argc, char *argv[]) { string inputS = argv[1]; string input[50]; /* Up to 50 tokens */ char buffer[200]; int i; int strnum = 0; char *next = buffer; char c; if (argc != 2) { fprintf(stderr, "Usage: %s expression\n", argv[0]); return 1; } printf("input: <<%s>>\n", inputS); printf("parsing:\n"); while ((c = *inputS++) != '\0') { input[strnum++] = next; if (isdigit(c)) { printf("Digit: %c\n", c); *next++ = c; while (isdigit(*inputS)) { c = *inputS++; printf("Digit: %c\n", c); *next++ = c; } *next++ = '\0'; } else { printf("Non-digit: %c\n", c); *next++ = c; *next++ = '\0'; } } printf("parsed:\n"); for (i = 0; i < strnum; i++) { printf("%d: <<%s>>\n", i, input[i]); } return 0; } 

Given that the program is called tokenizer and the command:

 tokenizer '(3+2)*564/((3+4)*2)' 

He gives me the conclusion:

 input: <<(3+2)*564/((3+4)*2)>> parsing: Non-digit: ( Digit: 3 Non-digit: + Digit: 2 Non-digit: ) Non-digit: * Digit: 5 Digit: 6 Digit: 4 Non-digit: / Non-digit: ( Non-digit: ( Digit: 3 Non-digit: + Digit: 4 Non-digit: ) Non-digit: * Digit: 2 Non-digit: ) parsed: 0: <<(>> 1: <<3>> 2: <<+>> 3: <<2>> 4: <<)>> 5: <<*>> 6: <<564>> 7: <</>> 8: <<(>> 9: <<(>> 10: <<3>> 11: <<+>> 12: <<4>> 13: <<)>> 14: <<*>> 15: <<2>> 16: <<)>> 

Answer 2

The simplest solution is to use a tool like flex to generate your lexer and let it do the job of breaking input into tokens (although flex expects its input to be received from a file stream, not an array of characters).

strtok() not a good solution for several reasons:

• It overwrites the input you might want to save for later use;
• It is a brute force tool and does not process a poorly formed inlet well;
• If you use your arithmetic operators as token separators, then the operators themselves will be knocked down.

The usual solution is to write the state machine (which is basically what it does for you). Here's a very quick n-dirty (focus on dirty) example:

 #include <stdio.h> #include <string.h> #include <stdlib.h> #include <ctype.h> /** * Read from a string specified by source, updating the pointer as we go. * We're assuming that token points to a buffer large enough to hold * our largest token; ideally, you would want to pass the length of the * target buffer and check against it, but I'm leaving it out for brevity. * * Tokens are either integers (strings of digits) or operators. * * Return 1 if we successfully read a token, 0 if we encountered an unexpected * character, and EOF if the next character is the end of the input string. */ int getToken(char **source, char *token) { enum {START, DIGIT, ERROR, DONE} state = START; size_t i = 0; char *operators="+-*/"; if (**source == 0) // at end of input return EOF; while (**source != 0) { switch(state) { /** * Initial state for this call. */ case START: if (isdigit(**source)) { state = DIGIT; token[i++] = *(*source)++; // append the digit to the token } else if (strchr(operators, **source) != NULL) { state = DONE; token[i++] = *(*source)++; // add the operator to the token token[i++] = 0; // and terminate the string } else if (isspace(**source)) { (*source)++; // ignore whitespace } else { /** * We've read something that isn't a digit, operator, or * whitespace; treating it as an error for now. */ state = ERR; } break; /** * We've read at least one digit. */ case DIGIT: if (isdigit(**source)) { token[i++] = *(*source)++; // append next digit to token } else { /** * We've read a non-digit character; terminate the token * and signal that we're done. */ token[i++] = 0; state = DONE; } break; case DONE: return 1; break; case ERR: return 0; break; } } return 1; } int main(int argc, char **argv) { char token[20]; char *input = argv[1]; for (;;) { int result = getToken(&input, token); if (result == 1) printf("%s\n", token); else if (result == 0) { printf("Bad character '%c'; skipping\n", *input); input++; } else if (result == EOF) { printf("done\n"); break; } } return 0; } 

Why (*source)++ instead of *source++ or source++ ? I don't want to update source , I want to update what source points to, so I need to dereference the pointer before applying ++ . The expression *(*source)++ basically means "give me the value of the character pointed to by the expression *source , and then update the value of *source ."

Answer 3

-> MAN STRCAT

 #include <stdio.h> #include <stdlib.h> #include <string.h> int main (int argc, const char **argv) { char *toto_str = "Toto"; char *is_str = "Is"; char *awesome_str = "Awesome"; char *final_str; size_t i; i = strlen(toto_str); i += strlen(is_str); i += strlen(awesome_str); final_str = malloc((i * sizeof(char)) + 1); strcat(final_str, toto_str); strcat(final_str, is_str); strcat(final_str, awesome_str); printf("%s", final_str); free(final_str); return 0; } 

Answer 4

strsep is a good choice here - grab the token and then decide what you want to do with it ...

char * string = "(3+ (5 + 6) / 8)"

char token; while ((token = strsep (& string, "(+ /)"))) {// Store the token ... if it is not (or) or a space}

Here - the token will be processed like Split () in Java / C #. This distorts the line when it is processed - however, with the correct delimiters - everything will be fine :)

Answer 5

It looks like you want to see the standard strtok .

Answer 6

this will give you an idea:

 #include <stdio.h> #include <string.h> main(int argc, char *argv[]) { printf("\nargv[1]: %s",argv[1]); char *p; p = strtok(argv[1],"+"); printf("\np: %s", p); p = strtok(NULL,"+"); printf("\np: %s", p); p = strtok(NULL,"+"); printf("\np: %s", p); printf("\n"); } 

This is just a sample code to demonstrate how this is done using the addition case only.
Get the basic idea of ​​this code and apply it in your code.
Sample output for this:

 ./a.out 5+3+9 argv[1]: 5+3+9 p: 5 p: 3 p: 9 

Again, I am only showing a + sign. You can check for p until it becomes NULL, and then perform the following operation, for example, subtraction, then multiplication, then division.