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When 'if' is issued from the Debugger prompt, always display the result

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This commit is contained in:
Jeff Parsons 2015-08-12 14:46:40 -07:00
commit 0a5267571f
1 changed files with 13 additions and 4 deletions
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17
modules/pcjs/lib/debugger.js
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@ -5152,19 +5152,28 @@ if (DEBUGGER) {
};
/**
* doIf(sCmd)
* doIf(sCmd, fQuiet)
*
* NOTE: Don't forget that the default base for all numeric constants is 16, so when you evaluate an
* expression like "a==10", it will compare the value of the variable "a" to 0x10; use a trailing period
* (eg, "10.") if you really intend decimal.
*
* Also, if no variable named "a" exists, "a" will evaluate to 0x0A, so the expression "a==10" becomes
* "0x0A==0x10" (false), whereas the expression "a==10." becomes "0x0A==0x0A" (true).
*
* @this {Debugger}
* @param {string} sCmd
* @param {boolean} [fQuiet]
* @return {boolean} true if expression is non-zero, false if zero (or undefined due to a parse error)
*/
Debugger.prototype.doIf = function(sCmd)
Debugger.prototype.doIf = function(sCmd, fQuiet)
{
sCmd = str.trim(sCmd);
if (!this.parseExpression(sCmd)) {
this.println("false condition: " + sCmd);
this.println("false: " + sCmd);
return false;
}
if (!fQuiet) this.println("true: " + sCmd);
return true;
};
@ -6372,7 +6381,7 @@ if (DEBUGGER) {
break;
case 'i':
if (asArgs[0] == "if") {
if (!this.doIf(sCmd.substr(2))) {
if (!this.doIf(sCmd.substr(2), fQuiet)) {
result = false;
}
break;