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CVE-2023-4504: OpenPrinting CUPS/libppd Postscript Parsing Heap Overflow

AHA! has discovered an issue with CUPS and libppd from OpenPrinting, and is publishing this disclosure in accordance with AHA!’s standard disclosure policy today, on Thursday, September 21, 2023. CVE-2023-4504 has been assigned to this issue.

Any questions about this disclosure should be directed to [email protected].

Executive Summary

Due to failure in validating the length provided by an attacker-crafted CUPS PPD file, CUPS version v2.5b1 and prior, by default, is susceptible to a heap-based buffer overflow, and possibly code execution. CVE-2023-4504 appears to be an instance of CWE-122, a heap-based buffer overflow.

Technical Details

The scan_ps function in the CUPS codebase provides functionality that scans through a string looking for the next Postscript object. When iterating through a string which contains an open parenthesis and ends with a single backslash (0x5c) character, the code incorrectly iterates forward a character without properly checking the bounds of the string resulting in a 1 byte read beyond the allocated heap buffer.

Snippet of the vulnerable code:

cups/cups/raster-interpret.c

1039 static _cups_ps_obj_t   *               /* O  - New object or NULL on EOF */
1040 scan_ps(_cups_ps_stack_t *st,           /* I  - Stack */
1041         char             **ptr)         /* IO - String pointer */
1042 {
...
1085   switch (*cur)
1086   {
1087     case '(' :                          /* (string) */
1088         obj.type = CUPS_PS_STRING;
1089         start    = cur;
1090
1091         for (cur ++, parens = 1, valptr = obj.value.string,
1092                  valend = obj.value.string + sizeof(obj.value.string) - 1;
1093              *cur;
1094              cur ++)
1095         {
1096           if (*cur == ')' && parens == 1)
1097             break;
1098
1099           if (*cur == '(')
1100             parens ++;
1101           else if (*cur == ')')
1102             parens --;
1103
1104           if (valptr >= valend)
1105           {
1106             *ptr = start;
1107
1108             return (NULL);
1109           }
1110
1111           if (*cur == '\\')
1112           {
1113            /*
1114             * Decode escaped character...
1115             */
1116
1117             cur ++;
1118
1119             if (*cur == 'b')
1120               *valptr++ = '\b';
1121             else if (*cur == 'f')
1122               *valptr++ = '\f';
1123             else if (*cur == 'n')
1124               *valptr++ = '\n';
1125             else if (*cur == 'r')
1126               *valptr++ = '\r';
1127             else if (*cur == 't')
1128               *valptr++ = '\t';
1129             else if (*cur >= '0' && *cur <= '7')
1130             {
1131               int ch = *cur - '0';
1132
1133               if (cur[1] >= '0' && cur[1] <= '7')
1134               {
1135                 cur ++;
1136                 ch = (ch << 3) + *cur - '0';
1137               }
1138
1139               if (cur[1] >= '0' && cur[1] <= '7')
1140               {
1141                 cur ++;
1142                 ch = (ch << 3) + *cur - '0';
1143               }
1144
1145               *valptr++ = (char)ch;
1146             }
1147             else if (*cur == '\r')
1148             {
1149               if (cur[1] == '\n')
1150                 cur ++;
1151             }
1152             else if (*cur != '\n')
1153               *valptr++ = *cur;
1154           }
1155           else
1156             *valptr++ = *cur;
1157         }

Line 1085 contains the case statement which provides the logic used to iterate through the given string.

On line 1091, the for loop within the case statement is used to iterate through each character after encountering an open paranthesis character (0x28), storing the pointer to the current character in cur.

On line 1111, the code checks if the current character is a backslash and finally, in line 1117, the character index is incremented without checking the length, now pointing to the null byte terminating the string.

Upon the next iteration of the loop, on line 1094, the loop now begins iterating through unallocated memory resulting in undefined behaviour.

A Base64 encoded blob of an example PPD PostScript command that can trigger the issue is below.

L///KFwAY3V1ZQ==


Attacker Value

By providing this malformed PPD file, an attacker could compromise the machine running the affected software, either CUPS or another application that uses the libppd library. Once compromised, this can provide an attacker a unique, privileged position in the targeted network.

Credit

This issue is being disclosed through the AHA! CNA and is credited to: zenofex and WanderingGlitch

Timeline


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