Borderou de evaluare (job #236305)
Utilizator |
|
IP | ascuns |
---|---|---|---|
Problemă | Control (clasa a 6-a) | Compilator | cpp |
Rundă | Arhiva de probleme | Status | evaluat |
Dată | 17 apr. 2016 13:41:08 | Scor | 0 |
Raport evaluator
Eroare de compilare:
user.cpp:1:19: error: too many decimal points in number
user.cpp:9:8: warning: multi-character character constant [-Wmultichar]
user.cpp:13:9: warning: multi-character character constant [-Wmultichar]
user.cpp:13:16: warning: multi-character character constant [-Wmultichar]
user.cpp:99:17: warning: character constant too long for its type [enabled by default]
user.cpp:1:1: error: ‘type’ does not name a type
user.cpp:2:1: error: ‘var’ does not name a type
user.cpp:3:5: error: ‘n’ does not name a type
user.cpp:4:6: error: found ‘:’ in nested-name-specifier, expected ‘::’
user.cpp:4:5: error: ‘g’ does not name a type
user.cpp:5:1: error: ‘procedure’ does not name a type
user.cpp:6:1: error: ‘var’ does not name a type
user.cpp:7:6: error: found ‘:’ in nested-name-specifier, expected ‘::’
user.cpp:7:5: error: ‘f’ does not name a type
user.cpp:8:1: error: ‘begin’ does not name a type
user.cpp:10:8: error: expected constructor, destructor, or type conversion before ‘(’ token
user.cpp:11:2: error: expected unqualified-id before ‘for’
user.cpp:14:10: error: expected constructor, destructor, or type conversion before ‘(’ token
user.cpp:15:2: error: ‘end’ does not name a type
user.cpp:16:1: error: ‘end’ does not name a type
user.cpp:18:1: error: ‘procedure’ does not name a type
user.cpp:19:1: error: ‘var’ does not name a type
user.cpp:20:7: error: found ‘:’ in nested-name-specifier, expected ‘::’
user.cpp:20:5: error: ‘ok’ does not name a type
user.cpp:21:1: error: ‘begin’ does not name a type
user.cpp:24:4: error: expected unqualified-id before ‘for’
user.cpp:28:10: error: ‘a’ does not name a type
user.cpp:29:10: error: ‘a’ does not name a type
user.cpp:30:10: error: ‘ok’ does not name a type
user.cpp:31:9: error: ‘end’ does not name a type
user.cpp:32:3: error: ‘until’ does not name a type
user.cpp:33:1: error: ‘end’ does not name a type
user.cpp:35:1: error: ‘procedure’ does not name a type
user.cpp:36:1: error: ‘var’ does not name a type
user.cpp:37:1: error: ‘begin’ does not name a type
user.cpp:38:7: error: ‘k’ does not name a type
user.cpp:38:12: error: ‘i’ does not name a type
user.cpp:38:17: error: ‘t’ does not name a type
user.cpp:39:2: error: expected unqualified-id before ‘while’
user.cpp:43:19: error: ‘i’ does not name a type
user.cpp:44:17: error: ‘end’ does not name a type
user.cpp:48:11: error: ‘k’ does not name a type
user.cpp:49:11: error: ‘j’ does not name a type
user.cpp:50:11: error: ‘i’ does not name a type
user.cpp:51:11: error: ‘t’ does not name a type
user.cpp:52:8: error: ‘end’ does not name a type
user.cpp:53:2: error: ‘b’ does not name a type
user.cpp:54:1: error: ‘end’ does not name a type
user.cpp:56:1: error: ‘procedure’ does not name a type
user.cpp:56:26: error: found ‘:’ in nested-name-specifier, expected ‘::’